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3 years ago

Find the angle in degrees

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

3 0

Answer:

90 degrees

Step-by-step explanation:

its a 45-45-90 triangle

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Part A

labwork [276]

An algebraic expression is one that consists of both number(s) and an alphabet(s). The required answers are:

i. Distance from Chenoa's house to the coffee shop = 6.0 miles

ii. Distance from coffee shop to Chenoa's school = 1.5 miles

iii. Distance from Chenoa's house to her school = 7.5 miles

An algebraic expression is one that consists of both number(s) and an alphabet(s). The alphabet is referred to as the unknown whose value has to be determined.

In the given question, let the distance from the coffee shop to Chenoa's school be represented by y.

So that;

The distance from Chenoa's house to the coffee shop = (2y + 3) miles.

The total distance from Chenoa's house to her school = 5y.

This implies that:

(2y + 3) + y = 5y

3y + 3 = 5y

3 = 5y - 3y

2y = 3

y = $\frac{3}{2}$

= 1.5

The distance from the coffee shop to Chenoa's school is 1.5 miles.

Thus;

(2y + 3) = ( 2(1.5) + 3)

= 6

The distance from Chenoa's house to the coffee shop is 6 miles.

And,

5y = 5(1.5)

= 7.5

The total distance from Chenoa's house to her school is 7.5 miles.

For more clarifications on algebraic expressions, visit: brainly.com/question/12792264

#SPJ1

5 0

2 years ago

Up to 12 people, p, can ride in the van.

Masteriza [31]

Answer:

12 people can fit so the van had enough sests

4 0

3 years ago

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.