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Firlakuza [10]
3 years ago
5

At a shelter, 15% of the dogs are puppies. there are 60 dogs at the shelter. How many are puppies?

Mathematics
2 answers:
frutty [35]3 years ago
8 0
So for this you need to find 15% of 60. There are a few ways to do this but I prefer to do it like this.

First, find 10% of 60. This can be found by dividing 60 by 10, because 100%/10% is 10.

This gives 10%=6

5% is half of 10%, so divide your answer for 10% (6), by 2.

This gives 5%=3

We needed 15%, so we add our 10% and 5% together, to get 9.

15% of 60 is 9, meaning there are 9 puppies.

Hope this helps!
Basile [38]3 years ago
7 0

Answer:

There~are~9 ~puppies.

Step-by-step explanation:

Multiply~by~a~fraction~to~find~15\%~of~60.

We~write~the~percentage~into~a~fraction. To~do~that~we~divide~it~by~100:

15\%~of~60~=\frac{15}{100}~of~60.

Now,~multiply:

\frac{15}{100}~of~60~=\frac{15}{100}  \times 60

                =\frac{900}{100}

                =9

15\%~of~60~is~9,~thus,~there~are~9~puppies.

Hope~this~helps!~and~have~an~amazing~day~ahead!

-Isabelle~Williams

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45% of the class is girls. There are 90 girls in the class. How many students are<br> in the class?
Stels [109]

Answer:

189

Step-by-step explanation:

45% x2= 90%

10% of 90= 9

90+90=180

plus that 10%

180 + 9 = 189

sorry if I'm wrong I'm trying

8 0
3 years ago
What does k equal if <br>485= 1/((k)(5.99x10^-3))
Verizon [17]
k\ne 0

485=\frac{1}{(k)(5.99*10^{-3})}\\\\ \frac{485}{1}= \frac{1}{0.00599k}\\\\ 485*0.00599k=1\\\\ k=\frac{1}{485*0.00599}\\\\k\approx 0.3442
7 0
2 years ago
Reflect order pair (-2, 5) over y=-x
vesna_86 [32]

When reflecting over y = -x

The x and y values change places and the signs change.

(-2,5) becomes (-5,2)

8 0
3 years ago
Read 2 more answers
X decreased by 7 is 38
marshall27 [118]

Answer:

45

Step-by-step explanation:

x - 7 = 38

add 7 to both sides to get the variable by itself

x = 45

7 0
3 years ago
F(x)=x-1/x^2-x-6 which is the graph of
hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

f\left(x\right)=x-\frac{1}{x^2}-x-6

x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:

\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0

-\frac{1}{x^2}-6=0

-1-6x^2=0

\mathrm{No\:Solution\:for}\:x\in \mathbb{R}

\mathrm{No\:x-axis\:interception\:points}

y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:

y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0

As we know that the domain of a function is the set of input or argument values for which the function is real and defined.

\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}

\mathrm{Since}\:x=0\:\mathrm{is\:not\:in\:domain}

\mathrm{No\:y-axis\:interception\:point}

\mathrm{Asymptotes\:of}\:-\frac{1}{x^2}-6:\quad \mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-6

\mathrm{Range\:of\:}-\frac{1}{x^2}-6:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)

The graph is attached below.

5 0
3 years ago
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