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2 years ago

Find the value of.(i) (3° + 4 -¹) × 2²

Mathematics

1 answer:

never [62]2 years ago

8 0

Answer:

Step-by-step explanation:

Using the rule of exponents

$a^{0}$ = 1 and $a^{-m}$ = $\frac{1}{a^{m} }$

Given

( $3^{0}$ + $4^{-1}$ ) × 2²

= (1 + $\frac{1}{4}$ ) × 4

= $\frac{5}{4}$ × 4

= 5

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Find the roots: x^2−12x+32=0. <br><br> Enter your answers in the boxes (numbers only).

AnnZ [28]

The answers are 4 and 8.

Because:

x^2 - 12x + 32 =

(x - 4)(x - 8)

x - 4 = 0

+4 +4

x = 4

x - 8 = 0

+8 +8

x = 8

8 0

3 years ago

(3xy2)3 simplified im not sure how to do this

Naddika [18.5K]

The answer is 9xy^2

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3 years ago

What is 9 - (-8) = 20 and how did you get that answer?

Colt1911 [192]

Answer: No solution

Step-by-step explanation:

9-(-8)=20 given

9+8=20 the two negatives cancel out, making the eight positive

17=20 this statement is not correct, so the answer is no solution

no solution

I'm not sure if there was supposed to be a variable in there somewhere, but I did my best to explain. Hope this helps!

4 0

3 years ago

Read 2 more answers

What is the ninth triangular number?

earnstyle [38]

27 I think..... Oh darn I'm replying to my self

5 0

3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$