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3 years ago

Which sums are equal to 1hold 6/8?

Mathematics

1 answer:

katovenus [111]3 years ago

6 0

A. = 7/8
b. = 7/4 = 1 3/4
c. = 7/4 = 1 3/4
d. = 7/4 = 1 3/4

= b,c,d

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Phoenix [80]

Answer:

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Step-by-step explanation:

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3 years ago

Find the general indefinite integral. (use c for the constant of integration. )

Vlada [557]

The indefinite integral will be $\int (7x^2+8x-2)dx=\dfrac{7x^3}{3}+4x^2-2x+C)$

<h3 /><h3>what is indefinite integral?</h3>

When we integrate any function without the limits then it will be an indefinite integral.

General Formulas and Concepts:

Integration Rule [Reverse Power Rule]:

$\int x^ndx=\dfrac{x^{n+1}}{n+1}}+C$

Integration Property [Multiplied Constant]:

$\int cf(x)dx=c\int f(x)dx$

Integration Property [Addition/Subtraction]:

$\int [f(x)\pmg(x)]dx=\int f(x)dx\pm \intg(x)dx$

[Integral] Rewrite [Integration Property - Addition/Subtraction]:

$\int (7x^2+8x-2)dx=\int 7x^2dx+\int 8xdx -\int 2dx$

[Integrals] Rewrite [Integration Property - Multiplied Constant]:

$\int (7x^2+8x-2)dx=7 \int x^2dx+ 8 \int xdx -2\int dx$

[Integrals] Reverse Power Rule:

$\int (7x^2+8x-2)dx= 7(\dfrac{x^3}{3})+8(\dfrac{x^2}{2})-2x+C$

Simplify:

$\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

So the indefinite integral will be $\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

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2 years ago

To solve for x in the equation x - 4 = 19, which inverse operation would you

lara31 [8.8K]

Add 4 to both sides......

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2 years ago

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Help? I cannot get my head around it.

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Answer:

60cm³

Step-by-step explanation:

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2 years ago

For the following exercises, solve each inequality and write the solution in interval notation.

Pie

Answer:

The solution of the given set in interval form is $(-\infty,-4] \cup[12, \infty)$ .

Step-by-step explanation:

It is given in the question an inequality as $|x-4| \geq 8$ .

It is required to determine the solution of the inequality.

To determine the solution of the inequality, solve the inequality $x-4 \geq 8$ and, $x-4 \leq-8$

Step 1 of 2

Solve the inequality $x-4 \geq 8$

$\begin{aligned}&x-4 \geq 8 \\&x-4+4 \geq 8+4 \\&x \geq 12\end{aligned}$

Solve the inequality $x-4 \leq-8$ .

$\begin{aligned}&x-4 \leq-8 \\&x-4+4 \leq-8+4 \\&x \leq-4\end{aligned}$

Step 2 of 2

The common solution from the above two solutions is x less than -4 and $x \geq 12$ .

The solution set in terms of interval is $(-\infty,-4] \cup[12, \infty)$ .

7 0

1 year ago