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3 years ago

The sum of two numbers(x and y) decreased by 76 is 38.

Mathematics

1 answer:

Fynjy0 [20]3 years ago

5 0

I will use w to represent the sum of x and y, and your answer is 114:

w - 76 = 38 (Add 76 over)

w = 114

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What is 19 x89532.089 x89 =<br><br><br> I need help please hurry

Zinaida [17]

151398762.5 that is the abswr

4 0

3 years ago

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Suppose Cone A is similar to Cone B and the scale factor between the solids is 3:2, respectively. If the height of Cone A is 15

blondinia [14]

Answer:

$10\ inches$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional

Let

x----> the height of cone B

using proportion

$\frac{3}{2}=\frac{15}{x} \\ \\x=2*15/3\\ \\x=10\ in$

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3 years ago

ratio of a rectangle's length to its width is 4:7. If the longer side is 31.5in., find the width, the perimeter, and the area of

Nikitich [7]

Answer:

Step-by-step explanation:

Ratio of length : width = 4 : 7

Length = 4x

Width = 7x

Longer side = 31.5 in

⇒ 7x = 31.5

Divide both sides by 7

x = 31.5/7

x = 4.5

length = 4x

= 4*4.5

l = 18 in

width = 31.5

w = 31.5 in

$\boxed{ \text {Perimeter of rectangle }= 2*l + 2*w}$

= 2*18 + 2*31.5

= 36 + 63

= 99 in

$\boxed{ \text{Area of rectangle = l*w}}$

= 18*31.5

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3 0

1 year ago

A model for the population in a small community after t years is given by P(t)=P0e^kt.

LUCKY_DIMON [66]

$\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}$

a)

so, if the population doubled in 5 years, that means t = 5. So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

$\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
I=\textit{initial amount}\to &1\\
r=rate\\
t=\textit{elapsed time}\to &5\\
\end{cases}
\\\\\\
2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r
\\\\\\
\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\
\textit{how long to tripling?}\quad 
\begin{cases}
A=3\\
I=1
\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}$

$\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t
\\\\\\
\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t$

b)

A = 10,000, t = 3

$\bf \begin{cases}
A=10000\\
t=3
\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I
\\\\\\
6597.53955 \approx I$

3 0

3 years ago

Can someone help me solve for b? 2(b + 3) = 12

Akimi4 [234]

2b+3=12
-3 -3
2b=9
B=4.5

7 0

3 years ago

Read 2 more answers