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Anna11 [10]
3 years ago
6

Mateo buys 5 postcards during each day of vacation. Write an equation that shows the

Mathematics
1 answer:
Over [174]3 years ago
8 0

Answer:

The answer is y=5x

Step-by-step explanation:

You might be interested in
Evaluate 1 plus negative 2/3 minus (-m) where m is 9/2
Mariana [72]

Answer:

Step-by-step explanation:

Negative 2/3 means -2/3

Hence;

1 + (-2/3) - (-m)

= 1 - 2/3 + m (positive x negative = negative, negative x negative = positive)

Substituting 9/2 for the value of m

= 1 - 2/3 + 9/2

The L.C.M = 6

= ( 6 - 4 + 27)/6

= 29/6

8 0
3 years ago
Why is it possible to have two different rational functions with the same asymptotes?
jolli1 [7]

Answer: Down below.

Step-by-step explanation:

A horizontal asymptote for a function is a horizontal line that the graph of the function approaches as x approaches ∞ (infinity) or -∞ (minus infinity). ... There are literally only two limits to look at, so that means there can only be at most two horizontal asymptotes for a given function

4 0
3 years ago
Kris has finished 7/8 of 40 questions. how many were done​
VLD [36.1K]

7/8=0.875

0.875 x 40= 35

Therefore, Kris has finished 35 questions.

Hope this helps.

8 0
2 years ago
Please help!! What is the equation of this line in slope-intercept form? <br><br>(see picture)​
faust18 [17]

Answer:

you answer is B

Step-by-step explanation:

I am almost 100% sure it is B but correct me if im wrong

5 0
2 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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