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nasty-shy [4]
3 years ago
7

Every weekday, Mr. Jones bikes from his home to his job. Sometimes he rides along two roads, the long route that is shown by the

solid lines. Other times, he takes the shortcut shown by the dashed line.
A right triangle. The distance from the angle with measure 90 degrees to Job is 15 kilometers. From the angle to Home is a. They hypotenuse is 17 kilometers.


How many fewer kilometers does Mr. Jones bike when he takes the shortcut instead of the long route?

6 kilometers

8 kilometers

23 kilometers

32 kilometers

Mark this and return

Mathematics
2 answers:
azamat3 years ago
8 0

Answer:

a

Step-by-step explanation:

on edge

notka56 [123]3 years ago
5 0

Answer:

(A)6 kilometers

Step-by-step explanation:

First, we determine the value of a using Pythagoras Theorem.

Hypotenuse^2=Opposite^2+Adjacent^2\\17^2=a^2+15^2\\a^2=17^2-15^2\\a^2=289-225\\a^2=64\\a^2=8^2\\a=8$ km

Therefore:

Distance along the long route = 8 + 15 =23 km

Distance along the shortcut =17 km

Difference =23-17 =6km

Therefore, Mr. Jones bikes 6km less when he takes the shortcut instead of the long route.

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Tpy6a [65]

Answer:

y intercept - (0, 275)

x intercept - (0, 125)

Step-by-step explanation:

5 0
3 years ago
Given the graph y = f(x), explain and contrast the effect of the constant c on the graphs y = f(cx) and y = cf(x).
Igoryamba
Given the graph y = f(x)

The graph y = f(cx), where c is a constant is refered to as horizontal stretch/compression

A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression is the squeezing of the graph towards the y-axis.  A compression is a stretch by a factor less than 1.

If | c | < 1 (a fraction between 0 and 1), then the graph is stretched horizontally by a factor of c units.
If | c | > 1, then the graph is compressed horizontally by a factor of c units.

For values of c that are negative, then the horizontal compression or horizontal stretching of the graph is followed by a reflection across the y-axis.


The graph y = cf(x), where c is a constant is referred to as a vertical stretching/compression.

A vertical streching is the stretching of the graph away from the x-axis. A vertical compression is the squeezing of the graph towards the x-axis. A compression is a stretch by a factor less than 1.

If | c | < 1 (a fraction between 0 and 1), then the graph is compressed vertically by a factor of c units.
If | c | > 1, then the graph is stretched vertically by a factor of c units.

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3 0
3 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
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Wewaii [24]

Answer:

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Firstly, we remove the parenthesis since there is nothing in front of it.

There is a “+” sign in front of the parenthesis, it stays the same.

Secondly, we combine like terms. We combine the x’s, the y’s, and the normal numbers without a variable.

Finally, after combining the like terms. You end up with “2x+3y+5”.

8 0
3 years ago
Can you please help me figure out this answer? Please and thank you
galben [10]

Here you're being asked to find the "perimeter" of the space, even tho' the problem doesn't specifically ask for it.

The formula for P is P = 2W + 2L.

Here the width, W, is 3 1/2 yds, and the length, L, is 4 2/3 yds. Subbing these two values into the formula for P (above) results in:

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= 7 yds + 9 1/3 yds = 16 1/3 yds, total.

7 0
3 years ago
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