In ΔOPQ, o = 1.2 inches, p = 3.2 inches and ∠Q=37°. Find the area of ΔOPQ, to the nearest 10th of a square inch.

Answer:

The value of the expression is 60.

Answer:

The two in the middle?

Answer:

2

Step-by-step explanation:

First substitute the values of z and y in the expression.

Use PEDMAS. We have to multiply 5 and 3, which gives 15. Subtract 1 from 15 and then divide the result by 7.

[tex]\sf \dfrac{5z + y}{7}= \dfrac{5*3 + (-1)}{7}\\\\[/tex]

          [tex]\sf = \dfrac{15 - 1}{7}\\\\ = \dfrac{14}{7}\\\\= 2[/tex]

The length and width of the rectangle is found as 19 inches and 5 inches respectively.

Now, as per the given question;

Let width of a rectangle = w inches

Then, length of a rectangle = 3w + 4

Perimeter of a rectangle = 2(length + width)

Substitute the values in the formula of perimeter.

48 = 2(3w + 4 + w)

48/2 = 4w + 4

24 = 4w + 4

20 = 4w

w = 5

Thus, the width of the rectangle is 5 inches.

Put w = 5 inches in its length.

l = 3w + 4

l = 3×5 + 4

l = 15 + 4

l = 19 inches.

Therefore, the length of rectangle is found as 19 inches.

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The ratio of the area of the smaller square to the area of the larger square = 4 : 49

Let the perimeter of the small square be [tex]P_1[/tex]

Lethe the perimeter of the large square be [tex]P_2[/tex]

Perimeter of a square = 4 Length

[tex]P_1=4L_1\\\\P_2=4L_2[/tex]

The ratio of the perimeters = 2:7

[tex]\frac{4L_1}{4L_2} =\frac{2}{7} \\\\\frac{L_1}{L_2} =\frac{2}{7}[/tex]

The area of a square = L^2

[tex](\frac{L_1}{L_2} )^2=(\frac{2}{7} )^2\\\\\frac{L_{1} ^{2} }{L_{2} ^{2}} =\frac{2^2}{7^2} \\\\\frac{L_{1} ^{2} }{L_{2} ^{2}} =\frac{4}{49}[/tex]

Therefore, the ratio of the area of the smaller square to the area of the larger square = 4 : 49

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tom has 8 cards and sarah has  24 cards .

tom and sarah total baseball cards =32

sarah has three times as many baseball cards as tom

The 2 equations are

1. sarah + tom = 32

2. sarah = [tex]3 \times tom[/tex]

substitute equation 2 in 1.

[tex]3\times tom +tom= 32[/tex]

4 tom = 32

tom = 32/4

tom = 8 cards

sarah = [tex]3\times tom[/tex]

         = [tex]3\times 8[/tex]

sarah = 24 cards

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Answer:

Step-by-step explanation:

a,d

Answer:ad

Step-by-step explanation:

ad

1.5 quarts of pure antifreeze must be added to 3 quarts of a ​40% antifreeze solution to obtain a 60​% antifreeze​ solution

Given;

Percentage of 3 quarts of antifreeze solution = 40%

We have to find,

Quarts to obtain the antifreeze solution of 60%

Now,

The 3 quarts of 40% antifreeze solution contains 0.4 x 3 = 1.2 quarts of pure antifreeze solute

Let,

x is the number of quarts of pure antifreeze needed to make a 60% solution

The equation to solve becomes, where x is the amount of pure antifreeze to be added is

(1.2 + x )/(3 + x) =0.6

(1.2 + x ) = 0.6 * (3 + x)

(1.2 + x ) = 1.8 + 0.6x

x-0.6x = 1.8 - 1.2

0.4x = 0.6

x = 1.5

Therefore,

1.5 quarts of pure antifreeze to the current solution to make a 60% solution

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Answer:

120 songs

Step-by-step explanation:

         Eva=4u

       Hank=1u

Eva+Hank=5u

            5u=600

             1u=600/5

                 =120

Answer: -12x

Step-by-step explanation: 4x +4 −2 −2x

Calculate x to the power of 1 and get x.

4x+4×( −2 −2x )

Calculate 2 to the power of −2 and get  41.

4x+4×( − 41x)

Divide x by − 41

​by multiplying x by the reciprocal of − 41 .

4x+4×( −1x×4)

Anything divided by -1 gives its opposite.

4x+4(−x×4)

Multiply 4 and −1 to get −4.

4x−4x×4

Multiply −4 and 4 to get −16.

4x−16x

Combine 4x and −16x to get −12x.

−12x

Answer:

Domain: 1 ≤ x ≤ 6

Range 10 ≤ M(x) ≤ 60

Step-by-step explanation:

M(x) = 10x where x represents the number of tickets bought

Assume nobody goes with Victoria. In that case she will go alone, only 1 ticket is bought and total cost = $10

Maximum 5 friends means total of 6 tickets purchased so total cost is $60

Domain refers to the possible input values ie possible x values. Here the domain is 1 ≤ x ≤ 6 also written as [1, 6]

The range is the total possible set of output values. Here output of the function is cost and cost ranges between 10 and 60 inclusive

So Range = 10 ≤ M(x) ≤60 also written as [10, 60]

Answer:

pretty sure its 2.9

Step-by-step explanation:

Answer:2.92 or just 3

Step-by-step explanation:

btw read your lesson and pay attentiom

Total 5040 passwords can be generated when no digits are repeated

• Permutation in mathematics is an arrangement of objects in an definite order. The elements or sets or things are arranged in sequential order or linear order.

• Permutation is classified into four types :

1) where repetition is not allowed

2) where repetition is allowed

3) objects that are non distinct  

4) circular permutation

• The formula for calculating permutation is n!/(n-r)!

As we are given that the password uses four digits from 0-9 and repetition is not allowed.

So, n = 10 and r = 4

Using the formula for permutation we get

= 10! / (10-4)!

= 10!/6!

= 10 * 9 * 8 * 7 * 6! / 6!

= 10 * 9 * 8 * 7

= 5040

Therefore, 5040 four digit passwords are possible when no digit is repeated.

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A measurement that can be repeated is said to be precise and one that is close to the accepted or actual value is approximate.

Given that,
The incomplete statement is given we have to complete the statement.

All over the world, most things are in use as of common quantity, so people all over the world agree that they have to follow a common system of measurement and that the common system of measurement is called standard measurement.

Here,

A measurement that can be repeated is said to be precise values
A measurement that is close to the accepted or actual value is approximate.

Thus, the required words are precise and approximate.

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The average speed of Lee is 70 km/hr.

To clear up this trouble we want to understand the concept of average speed.

average speed is given while we divide the overall distance blanketed with the aid of the total time taken.

right here, it's miles given that Lee and Johnson each drove a hundred and forty Km. additionally, it's miles given that lee drove for two hours.

So, average speed of Lee = [tex]\frac{140}{2}[/tex]

                                         = 70 km/hr

Here, the average speed at which Lee drove the car and went to business meeting is 70 km/hr.

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Answer:

[tex] \sf \large( \frac{2}{5} ) {}^{14} [/tex]

Step-by-step explanation:

=((2/5)^4(2/5)^3)^2

=((2/5^4+3))^2

=((2/5^7))^2)

=(2/5)^14

Using a geometric sequence, we have that:

Graph 1 models the situation.

The end behavior of the graph is given as follows:

As x decreases, y approaches the line y = 2.

A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q, hence the second term is the first multiplied by q, the third is the second multiplied by q, and so on.

For item 4, we have that:

This means that the first graph is correct.

The end behavior of a function is given by the limits of the function when x goes to infinity.

Hence, from the final graph, we have that:

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Answer:

79

Step-by-step explanation:

Let the numerator of the original fraction be 'x' and the denominator be 'y'. So, the original fraction becomes xy

According to the question,

Condition I,

If 4 is subtracted from the numerator of a fraction, its value becomes 1/3.

or,x−4y=13

or,3(x−4)=y

or,y=3x−12

Condition II,

If 5 is added to the denominator of the original fraction, the value becomes 1/2.

or,xy+5=12

or,2x=y+5

Put value of y from equation (i) in equation (ii), we get,

or,2x=(3x−12)+5

or,2x=3x−12+5

or,3x−2x−7=0

∴x=7

Put value of x in equation (i), we get,

or,y=3×7−12

or,y=21−12

∴y=9

So, (x,y) = (7,9)

Therefore, the required original fraction is 79.

Let f(x) be the integrand.

[tex]f(-2)=-8 \\ \\ f(0)=0 \\ \\ f(2)=8 \\ \\ f(4)=64 \\ \\ \implies \int^{4}_{-2} x^3 \text{ dx}=\approx \frac{1}{2}(2)[-8+64+2(0+8)] \\ \\ =\boxed{72}[/tex]

Answer:13

Step-by-step explanation:

Answer:

There are no values of f to make this equation true.

Hope this helps.

Step-by-step explanation:

In the second part of the equation, 6( 1/16f  - 3 ), 6 * 1/16f is 3/8f. If both sides subtract a 3/8f and both sides add 18, you get 18.5 = 0 meaning there are no values of f to make this equation true.

Shawn earn 11 per hour.

Hannah travel 25 miles in one gallon of gas.

A graph can be defined as a pictorial representation or a diagram that represents data or values in an organized manner.

given:

According to the graph the y- axis shows the earnings.

So, we have the coordinates are

(1, 11), (2, 22) and (3, 33)

So, Shawn earn = 22 - 11 / 2 -1 = 11 per hour.

Now, in second case

The coordinates are

(3, 75), (5, 125), (6, 150), (10, 250)

So, m= (125 - 75) / (5-3)

m= 50/2

m=25

So, Hannah travel 25 miles in one gallon of gas.

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Answer:

The answer is D.

Step-by-step explanation:

Answer:

in this case the answer is D

Step-by-step explanation:

your welcome

Given lines CD and EF are parallel. Therefore answer is (a)

How to find slope of  two lines and check whether they are parallel or perpendicular

Given are two points P (x₁, y₁) and Q ( x₂, y₂)

Then slope of the line  PQ is given by  (y₂-y₁)/(x₂-x₁)

If two lines have slopes m₁ and m₂ respectively

Then the lines are parallel when their slopes are equal i.e. m₁=m₂

and  the lines are perpendicular when m₁*m₂= -1

Given C(5,2), D(8,3),E(3,8) F(-6,5)

Then,

Slope of CD = (3-2)/(8-5)= 1/3

Slope of EF = (5-8) / (-6-3)  = 1/3

Slope of CD=Slope of EF

Therefore CD and EF are parallel

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The transformation that will preserve distance between points when applied to a figure in the (x, y) -coordinate plane is B . (x,y) → (-x+ 2, y + 7)

Let the coordinate be represented as

(x, y)

For a transformation to preserve distance between points when applied to a figure in the (x, y)-coordinate plane, the transformation must be any of the following:

From the list of options, we have

B . (x,y) → (-x+ 2, y + 7)

This transformation is a translation

Hence, the transformation that will preserve distance between points when applied to a figure in the (x, y) -coordinate plane is B . (x,y) → (-x+ 2, y + 7)

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A> The equation for the line is y = -0.1x + 2600

B> The value of the car is $16,500

From the table, we know that for every 6000 miles added there is a decrease in the value by $600.

A> So it is the function of one order

y = kx + b

25500 = 5000k + b

24800 = 12000k + b

After equating it we get

b = 26000 and k = -0.1

Now put the value of b and k in the equation, and we get

y = -0.1x + 26000

Therefore the equation is y = -0.1 + 26000

B> When we put the value of x = 95000, we get

y = -0.1 × 95000 + 26000

  = 16500

Therefore the cost of the car is $16,500

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Answer:

m=-13

Step-by-step explanation:

The average rate of change of function f(x)= [tex]\frac{1}{x+5}[/tex]  on the interval [1,1+h] is [tex]-\frac{1}{36+6h}[/tex]

Given,

The function = [tex]\frac{1}{x+5}[/tex]

We know the average rate of change of the function is [tex]\frac{f(b)-f(a)}{b-a}[/tex]

Where a and b are the intervals

a= 1

b=1+h

f(a) =f(1)= [tex]\frac{1}{1+5}[/tex]

=[tex]\frac{1}{6}[/tex]

f(b)= f(1+h)

=[tex]\frac{1}{1+h+5}[/tex]

=[tex]\frac{1}{6+h}[/tex]

f(b)-f(a)= [tex]\frac{1}{6+h}-\frac{1}{6}[/tex]

[tex]=\frac{6-(6+h)}{6(6+h)} \\=\frac{6-6-h}{36+6h}\\ =-\frac{h}{36+6h}[/tex]

Then,

[tex]\frac{f(b)-f(a)}{b-a} =\frac{-\frac{h}{36+6h} }{1+h-1}[/tex]

[tex]=\frac{-\frac{h}{36+6h} }{h} \\=-\frac{1}{36+6h}[/tex]

Hence, the average rate of change of function f(x)= [tex]\frac{1}{x+5}[/tex]  on the interval [1,1+h] is [tex]-\frac{1}{36+6h}[/tex]

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The shorter leg of the right triangle measures 3 units, while the larger lag measures 11 units.

Here we know that we have a right triangle, and we know that:The hypotenuse measures √130 units.

The perimeter measures 14 + √130 units.

If we define:

x = shorter leg.

y = larger leg.

Then, using the Pythagorean theorem, we can write:x^2 + y^2 = √130^2 = 130

And for the perimeter equation we know that:

perimeter = x + y + √130 = 14 + √130

We can simplify this equation to:x + y = 14Then we have the system of equations:

x^2 + y^2 = 130

x + y = 14

Isolating one variable in the second equation we get.

x = 14 - y

Replacing that in the other equation we get:

(14 - y)^2 + y^2 = 130

Now we can solve this for y.

196 - 28y + y^2 + y^2 = 130

2y^2 - 28y + 196 - 130 = 0

2y^2 - 28y + 66 = 0

y^2 - 14y + 33 = 0

Using Bhaskara's formula we can get:

y = (14 ± √( 14^2 - 4*1*33))/2*1y = (14 ± 8)/2

Now, remember that y is the larger leg of the triangle, then:y = (14 + 8)/2 = 11For the value of x we use:

x = 14 - y = 14 - 11 = 3x = 3the shorter leg of the right triangle measures 3 units, while the larger lag measures 11 units.

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