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ivanzaharov [21]
3 years ago
7

Find the missing number so that the equation has no solutions. X - 18 = 2x – 13

Mathematics
2 answers:
wel3 years ago
8 0

Answer:

X=-5

Step-by-step explanation:

gregori [183]3 years ago
4 0

Answer:

Step-by-step explanation: x-5 i think

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After 3 years, Remington earned $390 in simple interest from a CD into which he initially deposited $4000. What was the annual i
olganol [36]
R=I/pt
R=(390÷(4,000×3))×100
R=3.25%
3 0
3 years ago
Read 2 more answers
IM SO LOST PLEASE HELP
LiRa [457]
Check the picture below.

\bf \textit{lateral area of a cone}\\\\
LA=\pi r\stackrel{slant-height}{\sqrt{r^2+h^2}}~~
\begin{cases}
LA=539\pi \\
r=49
\end{cases}\implies 539\pi =\pi 49\sqrt{r^2+h^2}
\\\\\\
\cfrac{539\pi }{49\pi }=\sqrt{r^2+h^2}\implies 11=\sqrt{r^2+h^2}

3 0
3 years ago
Ms.ziesenis walked 1/4 of a mile in 1/2 of an hour. use multiplication to find the unit rate​
shusha [124]

Answer:

he would walk 1 mile in 2 hours

Step-by-step explanation:

which is very slow tbh

7 0
2 years ago
A line has a slope of -3 and y-intercept of 2.
Pie

AnswA line can be written in the form y = mx + b where m is the slope and b is the y intercept.

Since the slope is given as 4, the equation will be y = 4x + b

Plugging in the point (2,1) to the equation we get 1 = 4(2) + b or 1 = b + 8

Solving for b gives b = -7 so the equation will be y = 4x - 7er:

Step-by-step explanation:

5 0
3 years ago
Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
3 years ago
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