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3 years ago

Find the product by using suitable properties 15 x (-25) x (-4) x (-10)

Mathematics

1 answer:

Colt1911 [192]3 years ago

7 0

Answer:

−15000

Step-by-step explanation:

Solution:

15×(-25) = -375

(-4)×(-10)= -40

-375×-40= -15000

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The table represents a function.

mezya [45]

Answer:

-8

Step-by-step explanation:

You are given the following table representing the function f(x):

$\begin{array}{cc}x&f(x)\\-4&-2\\-1&5\\3&4\\5&-8\end{array}$

This means

$f(-4)=-2\\ \\f(-1)=5\\ \\f(3)=4\\ \\f(5)=-8$

Hence,

f(5)=-8

8 0

3 years ago

Read 2 more answers

In order to explain finicial concepts to your clients you need to have

Neporo4naja [7]

Im think its C. Accounting.

Hope that is correct.

7 0

3 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

Choose all values of a that make the following equation true: 3/4 = a/36.

Gekata [30.6K]

9514 1404 393

Answer:

B, D

Step-by-step explanation:

The equation can be solved by multiplying by 36.

3/4 = a/36

a = 36(3/4) = 108/4 = 27

The listed values that make the equation true are 27 and 108/4.

4 0

3 years ago

Please can someone help me with both questions. Thank you. Work is due tomorrow.

Mariulka [41]

So let's begin!
So the length is 25 and the entire perimeter is 106. Lets put these both of these numbers as information we know.

We know the length is 25.
We know the full perimeter is 106.
We also know a rectangle has 2 pairs of congruent sides.

Okay so, the length is 25.
Since there are two sides that are the same we would multiply 25 by 2 to get 50.

Now we know 2 sides out of the four sides on a rectangle.

106 is the final perimeter, if we subtract 50 from 106, we get 56. That is the measurement for both sides we're missing but we need to know the width for just one side.
So we divide 56 by 2 to get 28.
The width is 28.

Heres the show your work part:

25 x 2 = 50
Gets both lengths for rectangle.

106 - 50 = 56
Gets the product for the widths.

56 / 2 = 28
Gets the width of just one side.

Your final answer: 28

7 0

3 years ago