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3 years ago

Can someone help please? i don’t wanna fail :(

Mathematics

1 answer:

Shtirlitz [24]3 years ago

8 0

answer choice A could be correct

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Is 1/8 greater than 3

artcher [175]

No because 3 is a whole number and 1/8 is not

8 0

3 years ago

Passing through (-2,-6) and parallel to the line whose equation is y=-3x+4

Licemer1 [7]

Answer:

y = -3x - 12

Step-by-step explanation:

Lines that are parallel have the same gradient.

Therefore using the point (-2,-6) you can make your own equation.

-6 will be y, -2 will be x

All you have to do is find the y-intercept which is C in y=mx+c

-6 = -3(-2) + c

-6 = 6 + c

-12 = c

Therefore;

y = -3x - 12

6 0

2 years ago

100 points and brainliest

Black_prince [1.1K]

Answer: The answer is D

Step-by-step explanation: Hope this helps

6 0

3 years ago

Read 2 more answers

Integrate x+1/sqrt(x). I know that the answer is 2/3 • sqrt(x) • (x+3) + C, but I don't know how it is simplified from 2/3 • x^3

Andreas93 [3]

$\displaystyle\int\frac{x+1}{\sqrt x}\,\mathrm dx=\int\frac x{x^{1/2}}+\frac1{x^{1/2}}\,\mathrm dx$

$=\displaystyle\int x^{1/2}+x^{-1/2}\,\mathrm dx$

By the power rule,

$=\dfrac{x^{3/2}}{\frac32}+\dfrac{x^{1/2}}{\frac12}+C$

(this seems to be the step you're not getting?)

$=\dfrac23x^{3/2}+2x^{1/2}+C$

The next step is to pull out a common factor of $x^{1/2}$ from the antiderivative:

$x^{3/2}=x^{1/2+1}=x^{1/2}\cdot x^1$

so that the final result is

$\dfrac23x^{3/2}+2x^{1/2}+C=\dfrac23x^{1/2}(x+3)+C$

3 0

3 years ago

Find all constants a such that the vectors (a, 4) and (2, 5) are parallel.

Nezavi [6.7K]

Answer:

$\displaystyle a = \frac{8}{5}$ .

Step-by-step explanation:

Two vectors $(x_1,\, y_1)$ and $(x_2,\, y_2)$ are parallel to one another if and only if the ratio between their corresponding components are equal:

$\displaystyle \frac{x_1}{x_2} = \frac{y_1}{y_2}$ .

Equivalently:

$\displaystyle \frac{x_1}{y_1} = \frac{x_2}{y_2}$ .

For the two vectors in this equation to be parallel to one another:

$\displaystyle \frac{a}{4} = \frac{2}{5}$ .

Solve for $a$ :

$\displaystyle a = \frac{8}{5}$ .

$\displaystyle \frac{8}{5}$ would be the only valid value of $a$ ; no other value would satisfy the $\displaystyle \frac{x_1}{y_1} = \frac{x_2}{y_2}$ equation.

4 0

3 years ago