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2 years ago

Find the domain of the function (ƒ • g)(x) where ƒ(x) = 2x2 + x, g(x) = 5x – 1

1 answer:

7 0

Not sure if that's a multiplication or a composition. It doesn't matter. Both f and g have a domain of all real numbers. The domain of

$( f \circ g)(x)=f(g(x))$

is the same as the domain of g(x), all real numbers

Answer: all real numbers

You might be interested in

What's 9/5 divided by 14/15? im stuck

Varvara68 [4.7K]

Answer:

When you divide fractions, you multiply by the reciprocal, So, you would have 9/5 * 15/14 you can cancel, and get 9/1 * 3/14, and end up with 27/14,

5 0

2 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

2 years ago

Is 2/4 greater than 1/2

Triss [41]

A fraction with the same numerator (number on top) as another but with a smaller denominator (number on bottom) is a larger number. For example 1/2 is bigger than 1/3 which is bigger than 1/4, etc.

8 0

3 years ago

Read 2 more answers

Calcium is an essential nutrient for strong bones and for controlling blood pressure and heart beat. Because most of the body’s

irakobra [83]

Answer:

No, because the 95% confidence interval contains the hypothesized value of zero.

Step-by-step explanation:

Hello!

You have the information regarding two calcium supplements.

X₁: Calcium content of supplement 1

n₁= 12

X[bar]₁= 1000mg

S₁= 23 mg

X₂: Calcium content of supplement 2

n₂= 15

X[bar]₂= 1016mg

S₂= 24mg

It is known that X₁~N(μ₁; σ²₁), X₂~N(μ₂;δ²₂) and σ²₁=δ²₂=?

The claim is that both supplements have the same average calcium content:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

The confidence level and significance level are to be complementary, so if 1 - α: 0.95 then α:0.05

since these are two independent samples from normal populations and the population variances are equal, you have to use a pooled variance t-test to construct the interval:

[(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$t_{n_1+n_2-2;1-/2}= t_{25;0.975}= 2.060$

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} }= \sqrt{\frac{11*529+14*576}{12+15-2} } = 23.57$

[(1000-1016)±2.060*23.57* $\sqrt{\frac{1}{12} +\frac{1}{15} }$ ]

[-34.80;2.80] mg

The 95% CI contains the value under the null hypothesis: "zero", so the decision is to not reject the null hypothesis. Then using a 5% significance level you can conclude that there is no difference between the average calcium content of supplements 1 and 2.

I hope it helps!

3 0

2 years ago

The expression sin 57 is equal to

Basile [38]

Answer:

0.84

Step-by-step explanation:

The sine function is a trigonometric function.

We want to evaluate

$\sin(57 \degree)$

The sine function can be found on most scientific calculators.

Make sure your calculator is in degree and enter sin 57

It will give you 0.83867

If you correct to the nearest hundredth, you get:

0.84

3 0

3 years ago