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2 years ago

What is the width of a rectangle with an area of 5/8in2 and a length of 10inches

Mathematics

1 answer:

Aneli [31]2 years ago

6 0

Area = length * width
5/8 = 10 * width
width = (5/8) / 10
width = 1 / 16
width = 0.0625 inches

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What does x equal x + 3 ≤ –5 + 2x

Anna [14]

X+3≤ -5+2x

Add 5 to the other side. (the inverse of subtraction, since 5 is negative)

x+8≤2x

Subtract x on both sides.

8≤x <- the answer

I hope this helps!
~kaikers

8 0

3 years ago

If (x+2) is a factor of x^3-6x^2+kx+10, k=

Alja [10]

If a binomial $(x-a)$ is a factor of a polynomial, then $a$ is a root of this polynomial.

$(x+2)=(x-(-2)) \Rightarrow a=-2\\\\
(-2)^3-6\cdot(-2)^2+k\cdot(-2)+10=0\\
-8-24-2k+10=0\\
-2k=22\\\
\boxed{k=-11}
$

3 0

2 years ago

Write a quadratic function for each graph described.

user100 [1]

Answer:

$y=2x^2-\frac{4}{3}x-\frac{10}{3}$

Step-by-step explanation:

we know that

The roots of the quadratic function (x-intercepts) are

x=-1 and x=5/3

we can write the equation of the parabola as

$y=a(x+1)(x-\frac{5}{3})$

where

a is a coefficient

Remember that

The parabola pass through the point (5,40)

substitute the value of x and the value of y of the ordered pair in the quadratic equation and solve for a

x=5, y=40

$40=a(5+1)(5-\frac{5}{3})$

$40=a(6)(\frac{10}{3})$

$40=20a\\a=2$

substitute

$y=2(x+1)(x-\frac{5}{3})$

apply distributive property

$y=2(x^2-\frac{5}{3}x+x-\frac{5}{3})\\\\y=2(x^2-\frac{2}{3}x-\frac{5}{3})\\\\y=2x^2-\frac{4}{3}x-\frac{10}{3}$

see the attached figure to better understand the problem

6 0

2 years ago

The weight of a box varies directly as the volume of the box . If a 138 - pound box has a volume of 23 gallons , what is the wei

Marizza181 [45]

Answer:

150 lb

Step-by-step explanation:

138lb/23 gal = x/25 gal

cross multiply and solve for x

23x = 138·25

x = 3450/23

x = 150 lb

3 0

2 years ago

How many solutions does the system of linear equations have?

sattari [20]

There is no solutions to these linear equations. in order to find how many solutions there are, you first need to solve for the first variable in one of the equations, then substitute the result into the other equation. since both equations are almost the same and the same format, it won't have any solutions.

hope this helped, God bless!

3 0

3 years ago