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3 years ago

What is the slope of the line that passes through the poiņts (-10, 4) and (-10,8)?

Mathematics

1 answer:

anygoal [31]3 years ago

7 0

Answer:

Slope is undefined

Step-by-step explanation:

Slope is calculated by $\frac{y1-y2}{x1-x2}$ so now you put the points (-10,4) and (-10,8) into that.

Since both x values are the same you can tell that the line is a vertial line meaning the slope is undefined.

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Which of the following correctly expresses sin (70) + sin(30) as a product?

JulsSmile [24]

Answer:

2sin50 cos20

Step-by-step explanation:

We need to write sin (70) + sin(30) as a product. The formula used here is :

$\sin A+\sin B=2\sin (\dfrac{A+B}{2})\cos(\dfrac{a-b}{2})$

Here, A = 70 and B = 30

So,

$\sin 70+\sin 30=2\sin (\dfrac{70+30}{2})\cos(\dfrac{70-30}{2})\\\\\sin 70+\sin 30=2\sin 50\cos20$

So, the value of sin (70) + sin(30) is 2sin50 cos20. Hence, the correct option is (c).

7 0

3 years ago

HELP ASAP!!! Please give a step-by-step explanation, or you will be reported!

stepan [7]

For all the dogs: you have 2.65+2.46+3.67+2.91 = 11.69%

Now, the probability of getting a chihuahua would be 3.44/11.69=0.2942= 24.92

Step-by-step explanation:

there ya go!!

6 0

3 years ago

Read 2 more answers

PLEASE HELP<br> 3/4M -12 but m = 32

slava [35]

Answer:

Replace the variable m with 32

in the expression.

3/4⋅(32)−12 Simplify each term.

24−12 Subtract 12 from 24.

Step-by-step explanation:

4 0

3 years ago

How many cups are 125 grams

KengaRu [80]

Answer:

5/8 cup

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution

Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

$y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\$

for y=e^{6x}:

$y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\$

Now we replace in the differential equation y'' − y' − 30y = 0

for y=e^{-5x}:

$25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0$

for y=e^{6x}:

$36e^{6x}-6e^{6x}-30=0\\36-6+30=0$

Now, to know if both function are linearly independent we calculate the Wronskian

$W(f,g)=fg'-f'g$

$W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0$

I hope this is useful for you

Best regard

7 0

3 years ago