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3 years ago

What is (4/5)x (2/3)please help me out

Mathematics

1 answer:

Vikentia [17]3 years ago

3 0

Answer:

$\frac{8}{15}$

Step-by-step explanation:

$\frac{4}{5}$ × $\frac{2}{3}$ ( multiply values on numerators/ denominators together )

= $\frac{4(2)}{5(3)}$ = $\frac{8}{15}$

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Divide 1/2 and 3/4 and it can not be a mixed number.

mamaluj [8]

Answer:

Many options - 4/6 or 2/3 or 0.66

Step-by-step explanation:

When dividing fractions, we use KCF (keep, change, flip) so our new equation is 1/2 x 4/3. Just multiply! Your answer should be 4/6, and if it needs to be simplified, the answer is 2/3. The answer as a decimal is 0.66 !

4 0

3 years ago

Read 2 more answers

What is the surface area of a tennis ball with a diameter of 6.7?? With full work please.

Sedbober [7]

So the formula you have to do is A= (pie)(radius)(squared)

divide 6.7 by 2 to find the radius which is 3.35

now do 3.35 * 3.35 ( we are squaring it) which = 11.225

now 11.225 times PIE is 35.264377537

so the answer is 35.264377537

hope that helped!

6 0

3 years ago

Rewrite in the form . Then find q(x) and r(x).

gizmo_the_mogwai [7]

Answer:

q(x)= 2x+5

r(x)=6

b(x)=x+4

Step-by-step explanation:

I have found what is missing in your question: Rewrite 2x^2+13x+26/x+4 in the form q(x)+r(x)/b(x).

First, we have to divide the polynomial by x+4 and remember that is x times 2x.

By multiplying this we are going to get $2x^{2} + 8x$ and now we are doing separation of it and getting $2x^{2} + 13x + 26$ and getting $2x^{2} + 13x + 26/ x + 4 = 2x^{2} +8x/x+4=5x+26/x+4=2x + 5x+26/x+4$

Because 5 is showing x times 5 we are going to multiply (x + 4) with 5 and that is 5x + 20.

$5x+26/x+4 = 5x + 20/x+4 = 6/ x+4=5 + 6/x+4$

Now when we can get more simplified we are having:

$2x^{2} +13+26/x+4=2x + 5= 6/x+4$

3 0

3 years ago

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

3 0

3 years ago

The distance between the y value in the data and the y value predicted from the regression equation is known as the residual. wh

kupik [55]

A. SS = r²(SS_x)

B. SS = (1-r²)(SS_x)

C. SS = r²(SS_y)

D. SS= (1-r²)(SS_y)

The residual sum is given in the picture below so it is D which is correct.

3 0

3 years ago