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3 years ago

Let g(x)=5x-1 and h(x)=X^2-1 Solve: g(h(x))=74

Mathematics

1 answer:

AURORKA [14]3 years ago

4 0

Answer:

Let's replace the h(x) function in g(x) and then use 74 as a result on the axis y. The correct answer is 4 .

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Find the average rate of change of a function that contains the points (-2,3) and (2,5). -2, -1/2, 1/2, 2

UkoKoshka [18]

Answer:

The answer is " $\frac{1}{2}$ ".

Step-by-step explanation:

Given points:

$(-2,3)\\\\(2,5)\\\\$

average rate=?

$x_1=-2\\\\y_1=3\\\\x_2=2\\\\y_2=5$

$average\ rate=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{5-3}{2-(-2)}\\\\=\frac{5-3}{2+2}\\\\=\frac{2}{4}\\\\=\frac{1}{2}$

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3 years ago

This is hard my homework says, name three decimals between 16.4 and 16.5. Draw a number line estimating the placement of all fiv

babymother [125]

16.425
16.450
16.475

16.4_________16.425__________16.45__________16.475__________16.5

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3 years ago

If sinθ<0 and cosθ>0, then the terminal point determined by θ is in:

alukav5142 [94]

Answer:

D.) Quadrant 4

Step-by-step explanation:

Sin theta < 0 tells us the angle is in quadrant 3 or 4

The y component is negative

Cos theta >0 tells us the angle is in quadrant 1 or 4

The x component is positive

Therefore theta must be in quadrant 4

5 0

3 years ago

Evaluate the expression for the given values of the variables. Under selected conditions, a sports car gets 14 mpg in city drivi

Degger [83]

Answer:

28 gallons

Step-by-step explanation:

We are given that a sports car gets 14mpg in city driving and 19mpg for highway.

The model G= $\frac{1}{14} c+\frac{1}{19 }h$

Where G=Amount of gasoline used (in gal) for c miles driven in the city and h miles driven on the high way.

Amount of gas used in 14 miles car in the city driving=1 gal

Amount of gas used in 1 mile car in the city driving= $\frac{1}{14} gal$

Amount of gas used in c miles car in the city driving = $\frac{1}{14}c$ gal

Similarly, for car driving on highway

Amount of gas used in h miles for car driving on highway= $\frac{1}{19}h$

c=98 miles in the city

h=399 miles on the highway

We have to find the amount of gas used required to derive 98 miles in the city and 399 miles on the highway.

Substitute the value of c and h in the given expression

Then, G= $\frac{1}{14}\times 98+\frac{1}{19}\times 399=7+21=28 gal$

Hence, the amount of gas required to derive 98 miles in the city and 399 miles on the highway=28 gallons

6 0

3 years ago

Read 2 more answers

A closed rectangle box with a volume of 16 3 ft is made from two kinds of materials. The top and bottom are made of material cos

sashaice [31]

Answer:

The dimensions of the box are;

2 feet, by 2 feet, by 4 feet

Step-by-step explanation:

The given parameters are;

The volume of the closed box = 16 ft³

The material with which the top is made = The material with which the bottom is made

The cost of the material with which the top and bottom is made = 10 cents/(ft.²)

The cost of the material with which the side is made = 5 cents/(ft.²)

Let 'x', 'y', and 'z' represent the length, width, and height of the material, we have;

x·y·z = 16...(1)

The total cost of the box = 2×10×x×y + 2×5×x×z + 2×5×y×z

∴ The total cost of the box = 20·x·y + 10·x·z + 10y·z...(2)

From equation (1)

z = 16/(x·y)

Therefore, we have;

20·x·y + 10·x·z + 10y·z. = 20·x·y + 10·x·16/(x·y) + 10y·16/(x·y)

20·x·y + 160/(y) + 160/(x)

Differentiating, we get;

f'(x) = 20·y - 160/x²

f'(y) = 20·x - 160/y²

20·y - 160/x² = 0

y = 8/x²

20·x - 160/y² = 0

x = 8/y²

∴ x = 8/(8/x²)² = x⁴/8

8 = x⁴/x = x³

x = ∛8 = 2

The length, x = 2 feet

y = 8/x² = 8/2² = 2

The width, y = 2 feet

z = 16/(x·y) = 16/(2 × 2) = 4

The height of the box, z = 4 feet

Therefore, the dimensions of the box are 2 feet, by 2 feet, by 4 feet.

5 0

2 years ago