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2 years ago

9

Systems of equations Solve the system of equations. - 5x + 2y = 9 y = 7x y = 70

1 answer:

qwelly [4]2 years ago

4 0

Answer:

No soltion

Step-by-step explanation:

No soultuion

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What is the length of AC?

Blizzard [7]

Answer:

10 + 20 = 30

Step-by-step explanation:

If you subtract 5 from 25 you get 20 which is the length distance from b to c. Let's do the same to 15, 15 - 5 = 10. So now let's add our given lengths: 10 + 20 = 30.

4 0

2 years ago

Which expression has a value of 36?

IrinaK [193]

The answer is A or C

7 0

3 years ago

Use the distance formula, show that the points (4,0), (2,1), and (-1,-5) form the vertices of a right triangle

mario62 [17]

Step-by-step explanation:

The distance formula between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points.

$A(4,\ 0),\ B(2,\ 1),\ C(-1,\ -5)\\\\AB=\sqrt{(2-4)^2+(1-0)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4+1}=\sqrt5\\\\AC=\sqrt{(-1-4)^2+(-5-0)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}\\\\BC=\sqrt{(-1-2)^2+(-5-1)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45}$

If a ≤ b < c are the sides of the right triangle, then

a² + b² = c²

$\sqrt5$

used $(\sqrt{a})^2=a$ for a ≥ 0.

$AB^2+BC^2=AC^2$ therefore ΔABC is a right triangle.

6 0

2 years ago

If the Alpha company is 79% staffed and the Beta company is only 62% staffed, what is the relative change of staffing from the A

sweet [91]

Answer:

27.41%

Step-by-step explanation:

Data provided in the question

The staffed of Alpha company = 79%

The staffed of Beta company = 62%

Based on the above information, the relative change of staffing from Alpha to beta company is

As we know that

$\bold {\ Relative \ change = \frac{\alpha- \beta }{\beta }}$

$= \frac{(79 -62)}{62} \% \\\\= \frac{17}{62} \times 100\\\\=0.2741 \times 100\\\\=27.41\ \%$

By applying the above formula we can get the relative change and the same is to be applied so that the correct percentage could come

5 0

3 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago