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3 years ago

Can you please help me do this and explain how you did it so I can look at it for references thanks so much (:

Mathematics

1 answer:

arlik [135]3 years ago

5 0

Answer:

Step-by-step explanation:

By applying Pythagoras theorem in the given right triangles,

(Hypotenuse)² = (leg 1)² + (leg 2)²

a). (123.1)² = x² + (48.7)²

x² = 15153.61 - 2371.69

x² = 12781.92

x = 113.06 cm

b). (117)² = (82)² + x²

x² = 13689 - 6724

x = 83.46 mm

C). (117)² = t² + (10.3)²

t² = 13689 - 106.09

t² = 13582.91

t = 116.55 cm

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Tommy has 6 more than 3 times the amount of money in his bank account than Judy. Judy has $7500 more than a number in her accoun

Ad libitum [116K]

Answer:

Amount of money Judy has in her account is $\$(7500+n)$ .

Amount of money Tommy has in his account is $\$(22506+3n)$

Step-by-step explanation:

To find : Amount of money in Judy account and amount of money in Tommy account:

Solution:

Given:

Judy has $7500 more than a number in her account.

Let the number be 'n'.

So we can say that;

Amount of money in Judy account is equal to 7500 plus number.

framing in equation form we get;

Amount of money in Judy account = $\$(7500+n)$

Hence Amount of money Judy has in her account is $\$(7500+n)$ .

Now Given:

Tommy has 6 more than 3 times the amount of money in his bank account than Judy.

So we can say that;

Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.

framing in equation form we get;

Amount of money in Tommy account = $3(7500+n)+6 =22500+3n+6=\$(22506+3n)$

Hence Amount of money Tommy has in his account is $\$(22506+3n)$ .

8 0

3 years ago

-|-5| Is the answer greater than, less than, or equal to -5

ehidna [41]

Answer:

equal to

Step-by-step explanation:

3 0

2 years ago

Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C

Lesechka [4]

Answer:

Solution : $-\frac{3}{4}-\frac{3}{4}i$

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$