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2 years ago

Find the range of the function. f(x) = 10-x^2

Mathematics

1 answer:

BARSIC [14]2 years ago

7 0

Answer:
range = x ≤ 10
Step-by-step explanation:
Given in the question the function
f(x)=10-x^2
The range of a function is the complete set of all possible resulting values of the dependent variable F(x) after we have substituted the domain which is the values of x.
domain = set of all integers
range = x ≤ 10
f(x)=10-x²
since any negative value of x will be positive with squared so we will start from 0
f(0) = 10-0² = 10
f(1) = 10 - 1² = 9 = f(-1) = 10 -(-1)²
....and so on.

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The area of a square is tripled by adding 10 cm to one dimension and 12 cm to the other. Determine the side length of the square

Rina8888 [55]

Area = length *width
A = x²
3x² = (x+10)(x+12)
3x² = x²+10x+12x+120
3x² = x²+22x+120
3x²-3x² = x²+22x+120-3x²
0 = -2x²+22x+120
Factor to solve for x:
-2(x²-11x-60)
(x+4)(x-15)
x = -4 & x = 15 but we can't have a negative length so we eliminate 4. The length of the square is 15 cm.

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3 years ago

Jill says he did it wrong. He should have multiplied the bottom first (3 x 5) and then multiplied by the height. Explain to Jill

inna [77]

Answer:

Step-by-step explanation:

Jack's method is equivalent because of the "law of commutative property", which means that multiplication problems have the same result no matter the order.

2x5 is the same as 5x2

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3 years ago

9(2w−y)=21w−9y Solve for y

valentinak56 [21]

9(2w−y)=21w−9y

First you multiply the numbers in the parenthesis by 9

$18w-9y=21w-9y$

subtract. 21w-18w=3w

$-9y=3w-9y$

add. -9y+9y=0

$0=3w$

divide.

$0=w$

Final answer is 0.

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3 years ago

The legs if an isosceles triangle are (3x^2+4x+2) inches and the base is (4x-5) inches. Find the perimeter of the triangle

alexdok [17]

Answer:

$P=(6x^2+12x-1)\ in$

Step-by-step explanation:

we know that

An isosceles triangle has two equal sides and two equal angles

The perimeter of triangle is equal to

$P=2(3x^2+4x+2)+(4x-5)$

Eliminate parenthesis

$P=6x^2+8x+4+4x-5$

Combine like terms

$P=(6x^2+12x-1)\ in$

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3 years ago

In parallelogram DEFG, DH equals X +3, HF equals 3Y, GH equals 2X -5 and HE equals 5Y plus to find the values of X and Y

daser333 [38]

The values of X and Y are 30 and 11 respectively

<h3>How to determine the values of X and Y?</h3>

The figure that represents the complete question is added as an attachment

The given parameters are:

DH = X +3

HF = 3Y

GH = 2X -5

HE = 5Y

From the attached parallelogram, we have:

DH = HF

GH = HE

Substitute the known values in the above equation

X + 3 = 3Y

2X - 5 = 5Y

Make X the subject in X + 3 = 3Y

X = 3Y - 3

Substitute X = 3Y - 3 in 2X - 5 = 5Y

2(3Y - 3) - 5 = 5Y

Expand

6Y - 6 - 5 = 5Y

Evaluate the like terms

Y = 11

Substitute Y = 11 in X = 3Y - 3

X = 3*11 - 3

Evaluate

X = 30

Hence, the values of X and Y are 30 and 11 respectively