What is the area, in square units, of a triangle with vertices at (-1,-1), (3,-1), (2,2) PLEASE HELP

Mathematics

1 answer:

murzikaleks [220]2 years ago

5 0

Step-by-step explanation:

A=½bh

A=½(4•3)

A=½(12)

A=6 units sq

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Please explain the how you got the answers thanks asap

stich3 [128]

Answer:

a) 90 stamps

b) 108 stamps

c) 333 stamps

Step-by-step explanation:

Whenever you have ratios, just treat them like you would a fraction! For example, a ratio of 1:2 can also look like 1/2!

In this context, you have a ratio of 1:1.5 that represents the ratio of Canadian stamps to stamps from the rest of the world. You can set up two fractions and set them equal to each other in order to solve for the unknown number of Canadian stamps. 1/1.5 is representative of Canada/rest of world. So is x/135, because you are solving for the actual number of Canadian stamps and you already know how many stamps you have from the rest of the world. Set 1/1.5 equal to x/135, and solve for x by cross multiplying. You'll end up with 90.

Solve using the same method for the US! This will look like 1.2/1.5 = x/135. Solve for x, and get 108!

Now, simply add all your stamps together: 90 + 108 + 135. This gets you a total of 333 stamps!

5 0

2 years ago

A&b are complimentary angles a measures 32 what is the measure of b

tensa zangetsu [6.8K]

Answer:

B=58

Step-by-step explanation:

Complementary angles are angles where the sum equals 90°. So all you have to do is 90-32 and you get 58.

8 0

2 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$