All categories

4 years ago

SHORT ANSWER Use the Distributive

Mathematics

1 answer:

Monica [59]4 years ago

8 0

Answer:

5(5 + 2)

Step-by-step explanation:

The distributive property states that If 2 numbers have a common factor you can divide the common factor out. Similarily, if they share a common factor, you can distribute the common factor into each number in the parenthesis.

Find the factors of 25 & 10:

25: 1, 5, 25

10: 1, 2, 5, 10

Note that the largest common factor is 5. Divide 5 from both number:

(25 + 10)/5 = 5(5 + 2)

5(5 + 2) is your answer.

You might be interested in

Match each function name with its equation. <br> (Look at picture)

iragen [17]

Answer:

a. square root is y = $\sqrt{x}$

b. linear is y = x

c. cubic is $y = x^3$

d. quadratic is $y = x^2$

e. reciprocal squared is $y = \frac{1}{x^2}$

f. absolute value is y = |x|

g. reciprocal is $y = \frac{1}{x}$

h. cube root is $y = \sqrt[3]{x}$

Step-by-step explanation:

8 0

3 years ago

Can someone help me please

Leni [432]

80 because you find the area of the triangle and subtract it from the area of the rectangle to get 80

5 0

3 years ago

How do I solve this? I need help

Sedaia [141]

Answer: √51
—————————
a^2 + b^2 = c^2
a^2 + 7^2 = 10^2
a^2 + 49 = 100
a^2 = 51
√(a^2) = √51
a = √51

7 0

3 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

4 years ago

I need help with these

HACTEHA [7]

Answer:

7+4=11

Step-by-step explanation:

12/3 equalls 4 because the 12÷3=4.

7 0

3 years ago