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3 years ago

7

Isaac rides 3/4 miles. He then rides another 3 miles. How many miles did he ride?

1 answer:

Alex787 [66]3 years ago

5 0

Just add 3/4 to 3 so the answers is 3 3/4 miles

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In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm

jeyben [28]

Answer:

The length of the legs is 8.64cm and 14.64cm respectively

Step-by-step explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

$AB = 17$

$PAMC = 32$

$PBMC = 25$

From the attachment, we have:

$y + z = AB$

Since, M is the Midpoint

$y = z = \½AB$

Substitute 17 for AB

$y = z = \½ * 17$

$y = z = 8.5$

Also, from the attachment

$v + x + z = PAMC$

$v + x + y = 32$

Substitute 8.5 for y

$v + x + 8.5 = 32$

$v + x = 32 - 8.5$

$v + x = 23.5$ --------- (1)

Also, from the attachment

$v + w + z = 25$

Substitute 8.5 for z

$v + w + 8.5 = 25$

$v + w = 25 - 8.5$

$v + w = 17.5$ ----------- (2)

Subtract (2) from (1)

$v - v + x - w = 23.5 - 17.5$

$x - w = 6$

Make x the subject

$x = 6 + w$

Apply Pythagoras Theorem:

We have that:

$AB^2 = AC^2 + BC^2$

The above can be replaced with

$17^2 = x^2 + w^2$ (see attachment)

$289 = x^2 + w^2$

Substitute 6 + w for x

$289 = (6 + w)^2 + w^2$

$289 = 36 + 12w + w^2 + w^2$

$289 - 36 = 12w + 2w^2$

$253 = 12w + 2w^2$

Reorder

$2w^2 + 12w - 253 = 0$

Solve using quadratic equation:

$w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 2$

$b = 12$

$c = -253$

$w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}$

$w = \frac{-12 \± \sqrt{144 + 2024}}{4}$

$w = \frac{-12 \± \sqrt{2168}}{4}$

$w = \frac{-12 \± 46.56}{4}$

Split:

$w = \frac{-12 + 46.56}{4}$ or $w = \frac{-12 - 46.56}{4}$

$w = \frac{34.56}{4}$ or $w = \frac{-58.56}{4}$

$w = 8.64$ or $w = -14.64$

But length can't be negative

So:

$w = 8.64$

Recall that: $x = 6 + w$

$x = 6 + 8.64$

$x = 14.64$

<em>Hence, the length of the legs is 8.64cm and 14.64cm respectively</em>

5 0

3 years ago

What is the number of possible outcomes when tossing 5 coins at once?

drek231 [11]

<span>Each coin can be either heads or tails. The number of outcomes is equal to the amount of values the coins can take on (two) raised to the number of coins being tossed (five). Two times itself five times (2*2*2*2*2) equals 32 possible outcomes.</span>

3 0

4 years ago

Jessie has 160 rubber bracelets. She arranges the bracelets in piles of 40. Which model shows 160 ÷ 40?

Sergio [31]

Answer:

I am not sure what model you are talking about, no options to choose from

Step-by-step explanation:

You can divide both by 10, making the problem 16 / 4, which would look like 4 piles of four.

This means that the model that looks like 4 groups of 40

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3 years ago

Need help with this

Alecsey [184]

$\frac{x {}^{3} - 3x {}^{2} - 10x {}^{2} + 30x + 4x - 12 }{x - 3 } \\ \frac{x {}^{2} \times (x - 3) - 10x \times (x - 3) + 4 \times (x - 3) }{x - 3} \\ \frac{(x - 3)(x {}^{2} - 10x + 4 )}{x - 3} \\ x {}^{2} - 10x + 4$

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3 years ago

Help is appreciated. Question is worth 15 points.<br><br> Suppose that 2

GrogVix [38]

Answer:

huh

Step-by-step explanation:

5 0

3 years ago

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