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Natalija [7]
3 years ago
12

Please help me, i will mark the brainliest pls!

Mathematics
1 answer:
Burka [1]3 years ago
6 0
No because if you could use distributive property you would do 5*7 then 5*10 then you would get 35*50 which equals 1,750 and if you do it like it’s set up you would do 7*10 first cause PEMDAS so parentheses first so 7*10 is 70 then 70*5 is 350 and that’s not even close to the other answer so no you can’t .
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25. answer this question by using the graph and information​
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It would be B. 2.5 weeks

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8 0
4 years ago
5 + 20 and the multiplicative inverse of 5, write the product and then write the expression in standard form,
vladimir1956 [14]
X+4
Product is the answer to a multiplication.
The multiplicative inverse is the same as the reciprocal.
(5x+20) x 1/5= 5x+20/5
=x+4
4 0
3 years ago
CAN SOMEONE PLEASE HELP ME. I NEED HELP ON THIS QUESTION​
sleet_krkn [62]

Answer:

a segment cd only

Step-by-step explanation:

segment cd is the only line that passes through segment ab at a right angle

4 0
4 years ago
7 minus the quotient of 3 and p
olga_2 [115]
7-(3/p)
The quotient of 3 and p, is 3/p.
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3 0
4 years ago
A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, &amp; vertical shift of -1.
Simora [160]

Answer: \bold{y=\dfrac{1}{2}}

<u>Step-by-step explanation:</u>

f(x) = A sin (Bx - C) + D

  • amplitude = |A|
  • period =\dfrac{2\pi}{B}
  • phase shift =\dfrac{C}{B}
  • vertical shift = D

<u>A</u>

amplitude of 3 is given so  3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3

<u>B</u>

period of 6π is given so 6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}

<u>C</u>

\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C

<u>D</u>

vertical shift of -1 is given so -1 = D


Now, substitute the values of A, B, C, and D into the formula (above):

f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1


Next, solve when x = 2π

f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1

        = 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1

        = 3\ \bigg(\dfrac{1}{2}\bigg) - 1

        =\dfrac{3}{2}-\dfrac{2}{2}

        =\dfrac{1}{2}

6 0
3 years ago
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