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3 years ago

I need help graphing 2x-5y=5

Mathematics

1 answer:

viktelen [127]3 years ago

6 0

See attached picture for graph

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Pls help me I think it might be 14 but I’m not sure

mash [69]

Yes the answer is 14 good job

8 0

3 years ago

Solve tor r.<br> *a(5 – x) = bz - 8

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Assuming the given equation is

$a(5 - r) = bz - 8$

To solve for r, we first divide both sides by a.

$5 - r = \frac{ bz - 8}{a}$

We now subtract 5 from both sides:

$- r = \frac{ bz - 8}{a} - 5$

We now divide through by - 1 to get:

$r = 5 - \frac{ bz - 8}{a}$

$r = \frac{5a - bz + 8}{a} \: \: where \: a \ne0$

3 0

3 years ago

What is the standard form of the equation of the circle to the right?

Shkiper50 [21]

Answer:

$(y+2)^2+(x+3)^2=1$

Step-by-step explanation:

Because this circle has a center at (-3,-2) and a radius of 1, the equation is $(y+2)^2+(x+3)^2=1$ . Hope this helps!

3 0

3 years ago

A company has 2 machines that manufacture widgets. An older machine manufactures 25% defective widgets, while the new machine ma

Bingel [31]

Answer:

73.90%

Step-by-step explanation:

Let Event D=Defective, D' = Non Defective

Let Event N=New Machine, N' = Old Machine

From the given information:

$P(D|N')=0.25\\P(D|N)=0.09\\P(N)=0.7\\P(N')=0.3$

We are required to calculate the probability that a widget was manufactured by the new machine given that it is non defective.

i.e. $P(N|D')$

$P(D'|N')=1-P(D|N')=1-0.25=0.75\\P(D'|N)=1-P(D|N)=1-0.09=0.91$

Using Baye's Law of conditional Probability

$P(N|D')=\dfrac{P(D'|N)P(N)}{P(D'|N)P(N)+P(D'|N')P(N')} \\=\dfrac{0.91*0.7}{0.91*0.7+0.75*0.3}\\ =0.73897\\\approx 0.7390$

Therefore given that a selected widget is non-defective, the probability that it was manufactured by the new machine is 73.9%.

5 0

3 years ago

Question 19 of 28

Goryan [66]

Answer:

BC^2=10^2+30^2

Step-by-step explanation:

P=10
B=30

Using pythagorean theorem

$\\ \sf\longmapsto BC^2=10^2+30^2$

$\\ \sf\longmapsto BC^2=100+300$

$\\ \sf\longmapsto BC^2=400$

$\\ \sf\longmapsto BC=\sqrt{400}$

$\\ \sf\longmapsto BC=20$

8 0

3 years ago