3 years ago

A line has a slope of 5 and passes through the point

Mathematics

1 answer:

podryga [215]3 years ago

5 0

Answer:

Step-by-step explanation:

The equation is mx+b=y.

M=slope (in this case 5)

x=x coordinates

b=y intercept

y=y coordinates

For the x and y coordinates, I'll use (2,0).

5(2)+_=0

10+_=0.

In this case, we only have one awnser, -10. Since 10+(-10) are the only ways to equal 0.

Hope this helps,

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If the equation of a circle is (x - 2)2 + (y - 6)2 = 4, it passes through point (5, 6).

Salsk061 [2.6K]

Hello,

False

(5-2)²+(6-6)²=3²+0=9 and not 4

5 0

3 years ago

Read 2 more answers

NEED ANSWER ASAP REALLY IMPORTANT i will take any of the answers

dedylja [7]

$1. \: MC \: is \: tangent \: to \: (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC \: and \: MZ \: are \: tangent \: to \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM = \sqrt{ {BC}^{2} + {MC}^{2} } = \sqrt{ {5}^{2} + {12}^{2} } = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \: \tan(60°) = \frac{MC}{BC} = \frac{13 \sqrt{3} }{ BC}\Leftrightarrow BC = \frac{13 \sqrt{3} }{\tan(60°)} = \frac{13 \sqrt{3} }{ \sqrt{3} } = 13 \\ 5. \: MC = \sqrt{ {BM}^{2} - {BC}^{2} } = \sqrt{ {20}^{2} - {12}^{2} } = 16 \\ 6. \:BZ = \sqrt{ {BM}^{2} - {MZ}^{2} } = \sqrt{ {25}^{2} - {20}^{2} } = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30$