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Kryger [21]
4 years ago
6

Find the equation of the linear function

Mathematics
1 answer:
jonny [76]4 years ago
7 0

o.m.g it is so easy the answer is (B).L.O.L.

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The area of a square is 36 cm2. which represents the side length of the square?
Sindrei [870]
Let n be a side of the square. Then:
n²=36
n=√36=6 cm.
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5 0
3 years ago
Read 2 more answers
the base length of a particular triangle is 4 inches mote than its height. it the area of the triangle is 10.5 square inches, fi
IRINA_888 [86]

Answer:

<u>Height = 3 inches and Base length = 7 inches</u>

Step-by-step explanation:

1. Information provided to answer the question correctly:

Base of the triangle = x+ 4

Height of the triangle = x

Area of the triangle = 10.5 inches²

2. Find the base length and height

Let's recall the formula of the area of a triangle:

Area = (Base * Height)/2

Replacing with the values we know:

10.5 = x (x + 4)/2

21 = x² + 4x

x² + 4x - 21 = 0

(x + 7) (x - 3) = 0

(x₁ + 7) = 0 ⇒ x₁ = -7

(x₂ - 3) = 0 ⇒ x₂ = 3

We take  x₂ as the answer because x₁ = -7 is not a valid height.

<u>Height = 3 inches and Base length = 7 inches</u>

3 0
4 years ago
If a 20-foot tree casts a shadow of 15 ft how long is the shadow of a student who is 5 ft 4 in tall?​
irina1246 [14]

Answer:

4 feets

Step-by-step explanation:

Given that :

Height of tree = 20 feets

Shadow of tree = 15 ft

Height of student = 5ft 4 in ; 4 in = 0.3333 foot = 5.3333 feet

Shadow of student = s

Height of tree / shadow of tree = height of student / shadow of student

20 / 15 = 5.33333 / s

Cross multiply :

20 * s = 15 * 5.33333

s = 79.99995 / 20

s = 3.9999975

Shadow of student =, 4 feets

7 0
3 years ago
Part b at the airport to walk from the car to the waiting area by the gate was
Paha777 [63]

Answer:

24

Step-by-step explanation:

8 0
3 years ago
Write an equation in standard form for the line, where the points (-2, -1) and (0, 4) are on the line
DedPeter [7]

Answer:

An equation in standard form for the line is:

\frac{5}{2}x-y=-4

Step-by-step explanation:

Given the points

  • (-2, -1) and (0, 4)

The slope between two points

\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(-2,\:-1\right),\:\left(x_2,\:y_2\right)=\left(0,\:4\right)

m=\frac{4-\left(-1\right)}{0-\left(-2\right)}

m=\frac{5}{2}

Writing the equation in point-slope form

As the point-slope form of the line equation is defined by

y-y_1=m\left(x-x_1\right)

Putting the point (-2, -1) and the slope m=1 in the line equation

y-\left(-1\right)=\frac{5}{2}\left(x-\left(-2\right)\right)

y+1=\frac{5}{2}\left(x+2\right)

y=\frac{5}{2}x+4

Writing the equation in the standard form form

As we know that the equation in the standard form is

Ax+By=C

where x and y are variables and A, B and C are constants

so

y=\frac{5}{2}x+4

\frac{5}{2}x-y=-4

Therefore, an equation in standard form for the line is:

\frac{5}{2}x-y=-4

7 0
3 years ago
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