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padilas [110]
3 years ago
11

Which expression is the best estimate of the product of 7/8 and 8 1/10

Mathematics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

8

Step-by-step explanation:

i don't see options to select from but my guess would be 8 because 7/8 is very close to being one and 8 1/10 is close to 8;  8 x 1 = 8

You might be interested in
The residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 3 to 5. If there were 7
Zinaida [17]

Answer:

2967 yes votes

Step-by-step explanation:

Welcome to Brainly!

One way of doing this problem follows:

Find the ratio, r, mentioned in the problem statement.  To do this,

write the equation 3r + 5r = 7912 and solve this for r:

8r = 7912, so that r = 989.

Then the number of yes votes was 3(989), or 2967, and the number of no votes was 5(989), or 4947.

check:  Does the ratio 2967/4947 equal 3/5?  yes.

There were 2967 yes votes.

8 0
3 years ago
A cab company charges $7 boarding fee in addition to its meter which is $4 for every mile. The equation of the line that represe
skad [1K]

Answer:

A

Step-by-step explanation:

y=4x+7

3 0
3 years ago
ALEGBRA 1 ! help. me. please. this. is. so. hard. i dont. understand it.
ANEK [815]

Answer:

Part a) to find the maximum height of the snowball, you have to differentiate the function. Therefore you get ----> dh/dt= -32x-8 . Now equate this to zero and solve for x. x= (-1)/4 now sub this value in to find h(x) [note: i'm talking about the original function] . I got h max = h((-1)/4) = 9 which is the max height.

Step-by-step explanation:

I'm not too sure about the other questions. Sorry

8 0
3 years ago
For what value of a should you solve the system of elimination?
SIZIF [17.4K]
\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}

\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20
\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12

\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}

3a-10y=-8 \ \textgreater \  \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}
3a-10y-3a=-8-3a

\mathrm{Simplify} \ \textgreater \  -10y=-8-3a \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}-10
\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}

Simplify more.

\frac{-10y}{-10} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \  \frac{10y}{10}

\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \  y

-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{-8-3a}{-10}

\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \  -\frac{-3a-8}{10} \ \textgreater \  y=-\frac{-8-3a}{10}

\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \  6x+10\cdot \frac{8}{10-3a}=20

10\cdot \frac{8}{10-3a} \ \textgreater \  \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{8\cdot \:10}{10-3a}
\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \  \frac{80}{10-3a}

6x+\frac{80}{10-3a}=20 \ \textgreater \  \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}
6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}

\mathrm{Simplify} \ \textgreater \  6x=20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \  \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}

\frac{6x}{6} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \  x

\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{20-\frac{80}{-3a+10}}{6}

20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \  \frac{20}{1}-\frac{80}{-3a+10}

\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10

Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \  \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}
\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \  \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}

20\left(-3a+10\right)-80 \ \textgreater \  Rewrite \ \textgreater \  20+10-3a-4\cdot \:20

\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \  20\left(-3a+10-4\right) \ \textgreater \  Factor\;more

10-3a-4 \ \textgreater \  \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \  -3a+6 \ \textgreater \  Rewrite
-3a+2\cdot \:3

\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \  3\left(-a+2\right) \ \textgreater \  3\cdot \:20\left(-a+2\right) \ \textgreater \  Refine
60\left(-a+2\right)

\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \  \frac{10\left(-a+2\right)}{\left(-3a+10\right)}

\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \   \frac{10\left(-a+2\right)}{-3a+10}

Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}

Hope this helps!
7 0
3 years ago
Read 2 more answers
There are 4 white and 5 red (indistinguishable) balls in a bag. Suppose you draw one ballout at a time without replacement and s
Alchen [17]

Answer:

5/9

Step-by-step explanation:

What we have in this question are four white balls and 5 red balls.

This is the sample space

[(4w,0R) (4w,1R)(4w,2R)(4w,3R)(4w,4R)(0w,5R)(1w,5R)(2w,5R)

So we have 9 possible sets of events

The event number with the last ball being white = 5

Probability of the last ball drawn being white = 5/9

= 0.56

5 0
3 years ago
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