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Luba_88 [7]
3 years ago
5

How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?

Mathematics
2 answers:
Katen [24]3 years ago
8 0
One i think i’m not sure tho
Vsevolod [243]3 years ago
4 0
One is the correct answer
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Evaluate the expression for the given value of the variables.
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For every c substitute 4 and for every d substitute -2

c=4
d=-2

6c + 5d - 4c - 3d + 3c - 6d

= 6(4)+ 5(-2)- 4(4)- 3(-2)+ 3(4)- 6(-2)

=24+(-10)-16-(-6)+12-(-12)

=24-10-16+6+12+12

=28
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The segments shown below form a triangle
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13. Find the length of the diagonal of the rectangular prism.
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Step-by-step explanation:

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3 years ago
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Shkiper50 [21]
To find the inverse, we swap the variables y and x, then solve for the new y.

3a. y=\frac{3}{x-1}

Swapping the variables: x=\frac{3}{y-1}
Solving for y: x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}
The domain of this inverse is x ≠ 0.
3b. y=x^2-1

Swapping: x = y^2 - 1
Solving for y: y^2 = x + 1 \\ y = \sqrt{x+1}
The domain of this inverse is x ≥ -1.
3c. y=\sqrt[3]{\frac{x-7}{3}}
Swapping: x=\sqrt[3]{\frac{y-7}{3}}
Solving for y: x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7
The domain of this inverse is all real numbers.
4a. y=\frac{3}{x-1}, y=1+\frac{3}{x}
y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x
y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x

4c. y=\sqrt[3]{\frac{x-7}{3}}, y=3x^3+7
y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x
y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x



3 0
3 years ago
What is the remainder when the following polynomial division is performed? (y^4 – y^3 + 2y^2 + y – 1)/(y^3 + 1)
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Y4 - y3 + 2y2 + y - 1 over (y + 1) • (y2 - y + 1) <span>
</span>
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