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3 years ago

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3. Charles and Jaden share a reward of $63 in a ratio of 2 : 5. How much does Jaden get?

1 answer:

marissa [1.9K]3 years ago

3 0

Answer:

Jaden received: $45

Step-by-step explanation:

$63 : 7 = 9 => $9 * 5 = $45

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Aidan is participating in an 8 mile walk for his favorite charity. The organizers used a coordinate grid to plot the course. The

Travka [436]

Answer:

Aidan is 2 miles far from the ending point when he reaches the water station.

Step-by-step explanation:

The locations of the starting point, water station and ending point are (3, 1), (3, 7) and (3, 9), all expressed in miles. First we determine the distances between starting and ending points and between starting point and water station by the Pythagorean Theorem:

From starting point to ending point:

$D = \sqrt{(3-3)^{2}+(9-1)^{2}}$ (Eq. 1)

$D = 8\,mi$

From starting point to water station:

$d = \sqrt{(3-3)^{2}+(7-1)^{2}}$ (Eq. 2)

$d = 6\,mi$

The distance between the water station and the ending point is:

$s = D-d$ (Eq. 3)

$s = 8\,mi-6\,mi$

$s = 2\,mi$

Hence, Aidan is 2 miles far from the ending point when he reaches the water station.

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3 years ago

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Bingel [31]

Answer:

Yes.. you are correct!

Step-by-step explanation:

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2 years ago

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Irina18 [472]

Answer:

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3 years ago

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Which ratio is equivalent to 3/15

Rzqust [24]

Answer:

Step-by-step explanation:

The ratio of 6:30 would be equivalent to it.

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3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by <em>b</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since <em>a/b</em> < <em>c/d</em>, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by <em>b/d</em> to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and <em>a/b</em> < <em>c/d</em> tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

5 0

3 years ago