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2 years ago

I need helpppppppppppp

Mathematics

1 answer:

ratelena [41]2 years ago

7 0

The answer to question 3 is D. 15(4a+3). And the answer to question 4 is A. 90. Have a great day

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√3xy^3 • √8xy^4 = i need steps too.

liubo4ka [24]

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

2x²y exponit 7√6

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2 years ago

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8. Bob bought a pair of tennis shoes on sale

mr_godi [17]

Answer:

20%

Step-by-step explanation:

So, I would set this up as a proportion-

72 x

90 100

Cross multiply-

100 x 72 = 7200

Divide-

7200/90 = 80

Subtract-

100 - 80 = 20

So, it is a 20% decrease

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3 years ago

Select the graph that represents the correct answer to the equation. 5y = 3x - 5

Ivanshal [37]

Here is the graph, you can match it yourself:

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1 year ago

What is the area of the rectangle? shoe your work! do not round anything until the very end and round your final answer to the t

weeeeeb [17]

Answer:

The area of the rectangle is 42 units^2

Step-by-step explanation:

we know that

The area of rectangle is equal to

$A=LW$

In this problem

AB=DC

BC=AD

see the attached figure with letters to better understand the problem

we have that

$L=BC\\W=AB$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

$A(2,-1),B(5,2),C(12,-5),D(9,-8)$

Find out the distance BC

we have

$B(5,2),C(12,-5)$

substitute in the formula

$d=\sqrt{(-5-2)^{2}+(12-5)^{2}}$

$d=\sqrt{(-7)^{2}+(7)^{2}}$

$d_B_C=\sqrt{98}\ units$

Find out the distance AB

we have

$A(2,-1),B(5,2)$

substitute in the formula

$d=\sqrt{(2+1)^{2}+(5-2)^{2}}$

$d=\sqrt{(3)^{2}+(3)^{2}}$

$d_A_B=\sqrt{18}\ units$

Find out the area

$A=(L)(W)$

we have

$L=d_B_C=\sqrt{98}\ units$

$W=d_A_B=\sqrt{18}\ units$

substitute

$A=(\sqrt{98})(\sqrt{18})=42.0\ units^2$

4 0

3 years ago

What is partial quotients

natka813 [3]

Answer:

Is an equivalent part of a proper fraction, expressed in split parts with denominators whose LCM is the denominator of the fraction.

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2 years ago

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