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3 years ago

1. What type of association does this

Mathematics

2 answers:

mafiozo [28]3 years ago

5 0

Answer:

Step-by-step explanation:

LekaFEV [45]3 years ago

4 0

Answer:

Step-by-step explanation:

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( 7x + 1/2) + ( 7x - 7 and 1/2) Simplify

pishuonlain [190]

Answer:

The answer will be 7(2x - 1)

Step-by-step explanation:

1) Convert 7 and 1/2 to improper fraction. Use this rule: a b/c = ac + b/c

$7x + \frac{1}{2} + 7x - \frac{7 \times 2 + 1}{2}$

2) Simplify 7 × 2 to 14 .

$7x + \frac{1}{2} + 7x - \frac{14 + 1}{2}$

3) Simplify 14 + 1 to 15.

$7x + \frac{1}{2} + 7x - \frac{15}{2}$

4) Collect like terms.

$(7x + 7x) + ( \frac{1}{2} - \frac{15}{2} )$

5) simplify.

$14x - 7$

6) Factor out the common term 7.

$7(2x - 1)$

Therefor, the answer is 7( 2x - 1 ).

7 0

2 years ago

Need help ASAP Which number line correctly shows -3 - -1.5?

Alexandra [31]

Answer:

The top one I'm pretty sure.

3 0

3 years ago

Help meeeeeeeeeeeeeeee 1/2=2m m=

musickatia [10]

Answer:

.25

Step-by-step explanation:

5 0

2 years ago

Find B: 9.38 88 15.83 4

mario62 [17]

Answer:

by using Pythagoras law

h²=p²+b²

13²=9²+b²

169-81=b²

b=√88=2√22m is your answer

4 0

3 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago