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3 years ago

What are the answers to these thank you and please tell me how to get it

Mathematics

1 answer:

mote1985 [20]3 years ago

4 0

5-3(1/2x+2)=-7
-3(1/2x+2)=-7-5
(1/2x+2)=4
1/2x=2
x=4
just to give u an idea

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5. If 1/4 inch represents 5 yards, the number of inches needed to show a football field 100yards long is:

alexdok [17]

Answer:

The answer is 20/80.

Step-by-step explanation:

1/4=5 yards

? =100 yards

100÷5=20

1/4×20

=20/80.

PLEASE MARK ME AS THE BRAINIEST.

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2 years ago

Chelsea spent half of her weekly

Anvisha [2.4K]

Answer:

I believe it would be 16

Step-by-step explanation:

8 is half of 16 and if she had 8 dollars after spending half her allowance, she would've had 16 as her allowance

Final answer: Her allowance would be 16 dollars

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3 years ago

A painter earns $15 per hour.What is the minimum number of hours he must work to earn at least $200?

babunello [35]

12 hours in order to make 200 dollars

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3 years ago

Find an equation in standard form of the parabola passing through the points (1, -2), (2,-2), (3,-4)

Yuliya22 [10]

The standard form of a parabola is y=ax²+bx+c
use the three given points to find the three unknown constants a, b, and c:
-2=a+b+c............1
-2=4a+2b+c......... 2
-4=9a+3b+c...........3
equation 2 minus equation 1: 3a+b=0..........4
equation 3 minus equation 2: 5a+b=-2.........5
equation 5 minus equation 4: 2a=-2, so a=-1
plug a=-1 in equation 4: -3+b=0, so b=3
Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
the parabola is y=-x²+3x-4

double check: when x=1, y=-1+3-4=-2
when x=2, y=-4+6-4=-2
when x=3, y=-9+9-4=-4
Yes.

8 0

3 years ago

a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

slavikrds [6]

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

<u>It is assumed that weight of pyramid is proportional to its volume.</u>

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

<u>Since, base and height of small pyramid and original pyramid are in proportion.</u>

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

Learn more:

brainly.com/question/17950304

6 0

2 years ago

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