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likoan [24]
3 years ago
5

Please help will give brainliest!

Mathematics
2 answers:
Finger [1]3 years ago
7 0
1.5 ft is the answer
bonufazy [111]3 years ago
3 0

Answer:

1.5 ft

Step-by-step explanation:

Use the Pythagorean theorem and rearrange it to find the third length.

3.6^2 + x^2 = 3.9^2

3.9^2 - 3.6^2 = x^2

15.21 - 12.96 = 2.25

sqrt 2.25 = 1.5

x = 1.5

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timama [110]
I think it’s 0.1

1 pizza divided 10 slices = 0.1
4 0
3 years ago
Read 2 more answers
Rosetta drove her car 160 miles in 2 hours. Which of the following correctly expresses her rate as a ratio?
svp [43]

Rosetta drove her car 80 miles in one hour. Assuming that you ate talking about miles per hour you decide 160÷2=80 therefore, she drove 80 miles per hour (80/mph)

7 0
3 years ago
Read 2 more answers
Work out the percentage to change to decimal places when a price of 200 is increased to 210.99
olya-2409 [2.1K]

Answer:

105.495%

Step-by-step explanation:

210.99/200=1.05495

200 * 1.05495 = 210.99

6 0
3 years ago
*15 points easy question*
maw [93]

Answer:

\sqrt{13}

Step-by-step explanation:

We are given an isosceles triangle. We need to remember the important rule that the base angles are equal. In this question we need to find the value of 'x' using a perpendicular height of 3 and a base length of 4. In this question we can half the triangle in half to help us find the value of x. Also we need to use Pythagoras theorem to help us find 'x'. Pythagoras states that a² + b² = c² so we want to find the hypotenuse of the triangle. If we half the triangle we get 2 triangle both with a base length of 2 and a perpendicular height of 3 so,

⇒ State Pythagoras theorem

→ a² + b² = c²

⇒ Substitute in the values

→ 3² + 2² = c²

⇒ Simplify

→ 9 + 4 = c²

⇒ Simplify further

→ 13 = c²

⇒ Square root both sides to find the value of 'c'

→\sqrt{13}  =  c

The value of x is the square root of 13

8 0
3 years ago
9) p(a) = a) – 3a; Find p(-2)<br> A) 18<br> B) 198<br> C) -110<br> D) -2
Afina-wow [57]
ANSWER: B)198



Have a great day!
6 0
2 years ago
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