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3 years ago

Sarah walks to school, 1.5 miles in 30 minutes. what is the snswer to this

Mathematics

1 answer:

nevsk [136]3 years ago

5 0

Answer: 3 mi/h

Step-by-step explanation:

Here we have to find the speed $V$ at which Sarah walks to school, knowing the distance $d$ she has walked and the time $t$ she expends.

So, in this case we can use the following equation:

$V=\frac{d}{t}$

Where:

$d=1.5 mi$

$t=30 min \frac{1 h}{60 min}=0.5 h$

Solving:

$V=\frac{1.5 mi}{0.5 h}$

$V=3 mi/h$ This is Sarah's speed

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For what value of k, -4 is a zero of polynomial x2-x-(2k+2)?

bekas [8.4K]

Hmm
well, um
if -4 is a zero then if we subsitute -4 for x, we should get 0

so
0=(-4)²-(-4)-(2k+2)
0=16+4-2k-2
0=18-2k
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3 years ago

A box with no top is to be constructed from a piece of cardboard whose length measures 12 inches more than its width. The box is

MAXImum [283]

Answer:

Width is 12 inches and height is 4 inches.

Step-by-step explanation:

Let the width of the piece of cardboard be = w inches

Then length is = $w+12$ inches

As given, 4 inch squares are cut from each corner;

So, the width is = $w-8$ inches

Length is = $w+12-8=w+4$ inches

When we are folding the cut portion, that gives the height as = 4 inches

The volume is given by : length x width x height

And volume is given as 256 cubic inches

So, $256=(w+4)(w-8)(4)$

=> $256=(w^{2}-4w-32)4$

=> $64=(w^{2}-4w-32)$

=> $w^{2} -4w=96$

or $w^{2} -4w-96=0$

=> $w^{2} -12w+8w-96=0$

=> $w(w-12)+8(w-12)=0$

We get , $(w+8)=0$ ; w = -8(neglect the negative value)

$(w-12)=0$ ; w = 12 inches

And length = $12+12=24$ inches

Hence, dimensions are : L = 24 inches ; w = 12 inches and h = 4 inches

We can check by putting these values in equation : $256=(w+4)(w-8)(4)$

$256=(12+4)(12-8)(4)$

=> $256=(16)(4)(4)$

=> $256=256$

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3 years ago

2. Draw the image of RST under the dilation with scale factor 2/3 and center of dilation (1,-1). Label the image RST .

professor190 [17]

Answer:

From the graph: we have the coordinates of RST i.e,

R = (2,1) , S = (2,-2) , T = (-1,-2)

Also, it is given the scale factor $\frac{2}{3}$ and center of dilation C (1,-1)

The mapping rule for the center of dilation applied for the triangle as shown below:

$(x, y) \rightarrow (\frac{2}{3}(x-1)+1, \frac{2}{3}(y+1)-1)$

$(x, y) \rightarrow (\frac{2}{3}x -\frac{2}{3}+1 , \frac{2}{3}y+\frac{2}{3}-1)$

$(x, y) \rightarrow (\frac{2}{3}x+\frac{1}{3} , \frac{2}{3}y-\frac{1}{3} )$

Now,

for R = (2,1)

the image R' = $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(1)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{2}{3}-\frac{1}{3} )$

⇒ R' = $(\frac{5}{3} , \frac{1}{3})$

For S = (2, -2) ,

the image S'= $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ S' = $(\frac{5}{3} , -\frac{5}{3})$

and For T = (-1, -2)

The image T' = $(\frac{2}{3}(-1)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{-2}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ T' = $(\frac{-1}{3} , \frac{-5}{3})$

Now, label the image of RST on the graph as shown below in the attachment:

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2 years ago

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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