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Step2247 [10]
3 years ago
13

PLEASE HELPPP i’ll give brainliest

Mathematics
1 answer:
sashaice [31]3 years ago
3 0
The correct answer is the 3rd option !! good luck
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Solve the inequality.
MatroZZZ [7]
6x-4>8-2x

Add 2x on each side

8x-4>8

8x>12
x>12/8

x>3/2

The answer is B
4 0
3 years ago
Draw rectangles to show 3 1/2
Nataliya [291]
Just draw three rectangles and a half of one

5 0
3 years ago
PLSSSSS SOMEONE ANSWER THIS QUESTION
liq [111]

Answer:

12πx⁴, 15x⁷, 16x⁹

Step-by-step explanation:

Volume of a cylinder: πr²h

Volume of a rectangular prism: whl

Plugging in variables for the volume of a cylinder, we get: 3x²·(2x)²·π

3x²·(2x)² = 3·2·2·x·x·x·x

= 12·x⁴

=12x⁴

Now, we just multiply that by π.

12x⁴·π = 12x⁴π

A monomial is a 1-term polynomial, so 12x⁴π is a monomial.

Plugging in variables for the volume of a rectangular prism, we get: 5x³·3x²·x²

5x³·3x² = 5·3·x·x·x·x·x

= 15·x⁵

= 15x⁵

Now, we just multiply that by x².

15x⁵·x²

= 15·x·x·x·x·x·x·x

= 15·x⁷

=15x⁷

A monomial is a 1-term polynomial, so 15x⁷ is a monomial.

Same steps for the last shape, another rectangular prism:

2x²·2x³·4x⁴

2x²·2x³

= 2·2·x·x·x·x·x

= 4·x⁵

= 4x⁵

Now, we just multiply that by 4x⁴.

4x⁵·4x⁴

= 4·4·x·x·x·x·x·x·x·x·x·

= 16·x⁹

= 16x⁹

A monomial is a 1-term polynomial, so 16x⁹ is a monomial.

6 0
2 years ago
(MATH) (6) ((PHOTO))<br>label is ft​
arsen [322]

Answer:

6 feet

Step-by-step explanation:

Volume of a rectangular prism = L × W × H

15 feet × 8 feet × H = 720 feet ³

120H = 720

H = 720 ÷ 120

H = 6 feet

5 0
3 years ago
Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

7 0
3 years ago
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