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2 years ago

4(x+7) please help me on this problem

Mathematics

2 answers:

Anni [7]2 years ago

8 0

The answer would be 28x right?

tresset_1 [31]2 years ago

4 0

The answere is 4x+28
Hope I helped

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135° Solve for <2. 62 = [?] 45° &2 86° Enter

andrey2020 [161]

Answer:

$\displaystyle \angle2 = {49}^{ \circ}$

Step-by-step explanation:

remember that,

if a side of a triangle is extended then exterior angle so formed equal to the sum of the two opposite interior angles

thus our equation is

$\displaystyle \angle2 + {86}^{ \circ} = {135}^{ \circ}$

cancel out 86° from both sides:

$\displaystyle \angle2 = {49}^{ \circ}$

hence,

the measure of the missing angle is 49°

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2 years ago

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352, 345, 338, ... Find the 43rd term.

Artemon [7]

Answer:

Step-by-step explanation:

first term = a = 352

Common difference = d = second term - first term

= 345 - 352

= (-7)

nth term = a + (n-1)*d

43rd term = 352 + (43-1)*(-7)

= 352 + 42 *(-7)

= 352 - 294

= 58

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2 years ago

Mark runs 3/4 of a mile each day for 5 days. What is the total distance that Mark has run after 5 days? A.3 3/4 B.4 1/4 C.5 3/4

Evgen [1.6K]

3/4 x 5 = 3 3/4

Your answer is A.

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3 years ago

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Is 19%i already pass that it was easy

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2 years ago

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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

2 years ago