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Nataly [62]
3 years ago
7

Q- y = 7 is a line which is-(A) Parallel to y-axis (B) Parallel to x-axis (C) Passing through (7, 7) (D) Passing through origin​

Mathematics
1 answer:
siniylev [52]3 years ago
3 0

Answer:

(B) Parallel to x-axis

Step-by-step explanation:

y = 7

This is a constant line, in which the value of y is always of 7, no matter the value of x.

In a line in which the value of y is constant, this line is parallel to the x-axis, so the correct answer to this question is given by option b.

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Which number line shows the solution to fx-5 > 3?
sp2606 [1]

Answer:

8

Step-by-step explanation:

send <em>the</em><em> </em><em>-</em><em>5</em><em> </em><em>to</em><em> </em><em>where</em><em> </em><em>the</em><em> </em><em>+</em><em>3</em><em> </em><em>is</em><em> </em><em>then</em><em> </em><em>you</em><em> </em><em>can</em><em> </em><em>now</em><em> </em><em>add</em><em> </em><em>the</em><em> </em><em>value</em><em> </em><em>together</em><em>.</em>

7 0
3 years ago
drew makes $8 a week on allowance and pends $3. write an expression to show how much he has at the end of the week. Use parenthe
postnew [5]
X is how much he makes in a week. 7(8-3)=x
6 0
3 years ago
Help me? idk the answer :P
Alexus [3.1K]
\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\&#10;a^{-{ n}} \implies \cfrac{1}{a^{ n}}&#10;\qquad \qquad&#10;\cfrac{1}{a^{ n}}\implies a^{-{ n}}&#10;\qquad \qquad &#10;a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\&#10;-------------------------------\\\\&#10;\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}&#10;\\\\&#10;-------------------------------\\\\&#10;

\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\&#10;-------------------------------\\\\&#10;\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4

\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\&#10;-------------------------------\\\\&#10;\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}&#10;\\\\\\&#10;\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}

\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9
7 0
3 years ago
(HELP WITH THIS )Suppose ages of cars driven by company employees are normally distributed with a mean of 8 years and a
8090 [49]

Answer:

Let say 75% of  cars older than age  = Y  years

than 25 % cars of age less than Y Years

Z score corresponding to 25 %  =  -0.674

Z score = (Value - Mean) /SD

Value = Y

Mean = 8

SD = 3.2

Z score = -0.674

-0.674 = (Y - 8)/3.2

=> Y =5.84 Years

=> Y ≈ 5.9 Years

Approximately 75% of cars driven by company employees are older than  about 5.9 years

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4 0
3 years ago
Help meeeeeeeeeeee:(
Vaselesa [24]

Answer:

1/3

Step-by-step explanation:

(1/3)^2x(1/3)^-1

so, we can write this as:

1/3 x 1/3 x 3/1

if calculated by cutting, the answer is 1/3

There is also an easier way to do this,

(1/3)^2 x (1/3)^-1

as the bases are same, the powers will be added.

2+(-1)

2-1

=1

so,

(1/3)^1 or 1/3

7 0
3 years ago
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