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RSB [31]
3 years ago
5

PLEASE HELP. !!!!!!!!!

Mathematics
1 answer:
madam [21]3 years ago
8 0

Answer:

Step-by-step explanation:

The formula for this is:

6(6 + x) = 3(3 + 25) and

6(6 + x) = 3(28) and

36 + 6x = 84 and

6x = 48 so

x = 8

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12 cm
Ahat [919]

Answer:

The answer is 200 cm³

Step-by-step explanation:

4 0
3 years ago
The points (-4,-4), (-4,4), (4,4), (4,-4) form a square on a coordinate plane. How long is a side length of a square?
Crank
The answer is 4 units long and 8 units wide
6 0
4 years ago
Read 2 more answers
Help needed!! Don’t know how to do!!
Aleks [24]

Answer:

D,0,2,-2

Step-by-step explanation:

2x^5-3x^3-20x=0

x(2x^4-3x^2-20)=0

x=0

or 2x^4-3x^2-20=0

put x²=t

2t²-3t-20=0

-20×2=-40

8-5=3

8×-5=-40

2t²-(8-5)t-20=0

2t²-8t+5t-20=0

2t(t-4)+5(t-4)=0

(t-4)(2t+5)=0

t=4

x²=4

x=2,-2

t=-5/2

x²=-5/2

it gives imaginary  root. so real rational roots are 0,2,-2

3 0
3 years ago
Read 2 more answers
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
zloy xaker [14]

Answer:

t=-20ln\left(\dfrac{1}{3}\right)

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

A=45e^{-0.05t}

We want to determine the value of t for which the area, A(t)=15 square kilometers.

15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0
3 years ago
SOMEONE PLEASE HELP!! I've tried this question twice and I just can't seem to get it
jolli1 [7]

\underline{\underline{\large\bf{Given:-}}}

\red{\leadsto}\:\textsf{Total Number of fish }\sf = 30

\red{\leadsto}\:\textsf{ Number of carp fish }\sf = 9

\underline{\underline{\large\bf{To Find:-}}}

\orange{\leadsto}\:\textsf{ Ratio of carp to sticklebacks to tench }\sf in  \:the  \:pond

\\

\underline{\underline{\large\bf{Solution:-}}}\\

\longrightarrowNumber of fish that are not carp = Total fish - number of carp fish

\begin{gathered}\implies\quad \sf 30-9 \\\end{gathered}

\begin{gathered}\implies\quad \sf 21 \\\end{gathered}

\longrightarrowSince 1/3 of remaining fish which are not carp are sticklebacks , Number of sticklebacks =

\begin{gathered}\implies\quad \sf  \frac{1}{3} \times number  \:of  \:remaining \: fish\\\end{gathered}

\begin{gathered}\implies\quad \sf  \frac{1}{3}\times 21 \\\end{gathered}

\begin{gathered}\implies\quad \sf  7 \\\end{gathered}

<u>Now</u> ,

\longrightarrowNumber of tench = Total number of fish - (Number of carp + Number of sticklebacks)

\begin{gathered}\implies\quad \sf  30-(9+7) \\\end{gathered}

\begin{gathered}\implies\quad \sf  30-16\\\end{gathered}

\begin{gathered}\implies\quad \sf  14\\\end{gathered}

\thereforeRatio of carp to sticklebacks to tench in the pond =

\begin{gathered}\quad\quad \boxed{\sf {9: 7:14}} \\\end{gathered}

3 0
2 years ago
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