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3 years ago

Y+5=-11 What does y equal?

Mathematics

1 answer:

Brut [27]3 years ago

8 0

Y+5=-11
-5 -5
Y= -16

Answer:-16

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A semi-circle is connected to an isosceles triangle. The radius of the semi-circle is 2 inches, and the height of the triangle i

NikAS [45]

Answer:

Step-by-step explanation:

Is there like a multiple choice so I could answer it:)

5 0

3 years ago

if the velocity of a body changes from 13m/s to 30m/s while undergoing constant acceleration whats the average velocity of the b

34kurt

Average speed = (1/2) (beginning speed + ending speed)
= (1/2) ( 13 m/s + 30 m/s )
= (1/2) ( 43 m/s )
= 21.5 m/s

3 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP THANK YOU<br><br>1. Find f(2) if f(x) = 2x + 1

Olin [163]

Answer:

F(2)?

Step-by-step explanation:

4 0

3 years ago

64 POINT QUESTION<br><br> √(5-2x) - √(2-x) = 1

Ivanshal [37]

Answer:

x = 2 and x = -2

Step-by-step explanation:

The easiest way is to guess. We know that you can't have a square root of a negative number. This constrains x between negative infinity to 2. Testing 2 and -2, we see that it satisfy the equation.

The more systematic approach is shown below

$\sqrt{5-2x}-\sqrt{2-x}=1 \\\sqrt{5-2x}-\sqrt{2-x}+\sqrt{2-x}=1+\sqrt{2-x} \\\sqrt{5-2x} = 1+\sqrt{2-x}$

$5-2x=(1+\sqrt{2-x})^2\\5-2x=1+2\sqrt{2-x}+2-x\\2-x=2\sqrt{2-x}\\1-0.5x=\sqrt{2-x}\\(1-0.5x)^2=2-x\\1-x+0.25x^2=2-x\\0.25x^2-1=0\\0.25x^2 = 1\\x^2 = 4\\x = +/-2$

x= -2, 2

7 0

3 years ago

A triangle has vertices at (-2,-3),(4, -3), and (3,5). What is the area of the triangle?

Bezzdna [24]

Answer:

this triangle is ABC with A(-2 ; -3), B(4; - 3) and C(3;5)

=> $AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} } }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2} }=\sqrt{5^{2}+8^{2} }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2} }=\sqrt{65}$

using Heron theorem, we have:

$S=\sqrt{p(p-AB)(p-AC)(p-BC)}\\\\S=\sqrt{(\frac{6+\sqrt{89}+\sqrt{65} }{2})(\frac{\sqrt{89}+\sqrt{65}-6 }{2} )(\frac{6+\sqrt{65}-\sqrt{89} }{2})(\frac{6+\sqrt{89}-\sqrt{65} }{2}) }\\\\S=24 \\\\$

with S is the area of the triangle

$p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65} }{2}$

Step-by-step explanation:

5 0

3 years ago