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lions [1.4K]
3 years ago
12

I need help if y’all want y’all can help me I need help

Mathematics
1 answer:
Snezhnost [94]3 years ago
6 0

The answer is D. 42

A triangle equals 180 in total so what you would do it subtract 122 from 180 because 180 is also the angle of a straight line. You would get 58 then you add 58 and 80 to get 138. Subtract 138 from 180 and you get your answer of 42

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What is the equation of the line parallel to the given line with an X intercept of 4
mixas84 [53]

You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.


My invented line: y = (2/3)x + 3


The new line MUST have the same slope: m = 2/3.


Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:


-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is


y = (2/3)x - 8/3.

3 0
4 years ago
Read 2 more answers
What is the 214 term in the sequence 7,10,13,16?
mixer [17]
We can figure this out using the explicit formula.

f(n)=f(1)+d(n-1)

n represents the term we are looking for.
f(1) represents the first term in the sequence, which in this case, is 7.
d represents the common difference, which in this case, is +3.

f(n) = 7 + 3(n - 1)
f(n) = 7 + 3n - 3
f(n) = 4 + 3n

Now, we can input 214 for n and solve.

f(214) = 4 + 3(214)
f(214) = 4 + 642
f(214) = 646

The 214th term in this sequence is 646.
5 0
4 years ago
PLEASE HELP FAST!!!!!
Anit [1.1K]
The answer to the first question is 7(3.14)/6.

the answer to the second question is quadrant 2
8 0
4 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
Tell me please we are timed hurry
padilas [110]

Answer: 255

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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