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3 years ago

7

CAN A TRIANGLE HAVE 2 RIGHT ANGLES ? EXPLAIN.

1 answer:

goldenfox [79]3 years ago

8 0

<span>So the question is can a triangle have two right angles. In Euclidian (flat) geometry the sum of all angles, in a triangle, by definition, must be equal to 180 degrees. Right angle, again, by definition, is an angle which has 90 degrees. So if we had two right anges in a triangle it wouldnt be a triangle anymore. The sides wouldn't connect. So, we can't have two right angles in a triangle, only one, because the sum of two other angles must be 90 degrees. </span>

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Salsk061 [2.6K]

Answer:

22

Step-by-step explanation:

f. 24-(5-9+18)÷7

24-(14)÷ 7

24 - 2 = 22

hope it helps.....

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3 years ago

Read 2 more answers

Mira is sending a care package to her cousin overseas. she wants to send a 64 ounce container filled with her cousins two favori

aalyn [17]

Answer:

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Step-by-step explanation:

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3 years ago

The polynomial p(x)=3x^3-20x^2+37x-20 has a known factor of (x-4). Rewrite p(x) as a product of linear factors. p(x) =

vesna_86 [32]

Answer:

$p(x) = (3x^{2} - 8x + 5)(x - 4)$

Step-by-step explanation:

p(x) has degree 3

(x - 4) has degree 1

So p(x) can be rewritten as a polynomial of degree 2 multiplying (x-4)

So

$p(x) = (ax^{2} + bx + c)(x - 4)$

To find a, b and c, we have to divide p(x) by x - 4.

So

Finding a:

Dividing the first term of p(x) by x - 4.

$\frac{3x^{3}}{x - 4} = 3x^{2}$

So a = 3.

Now multiplying 3x² by, x - 4, we have:

$3x^{2}(x - 4) = 3x^{3} - 12x^{2}$

Subtracting p(x) from this:

$3x^{3} - 20x^{2} + 37x - 20 - (3x^{3} - 12x^{2}) = 3x^{3} - 20x^{2} + 37x - 20 - 3x^{3} + 12x^{2} = -8x^{2} + 37x - 20$

Finding b:

Dividing the first term, after the subtraction, by x - 4.

$\frac{-8x^{2}}{x - 4} = -8x$

So b = -8.

Multiplying -8x by x - 4, we have:

$-8x(x-4) = -8x^{2} + 32x$

Then

$-8x^{2} + 37x - 20 - (-8x^{2} + 32x) = -8x^{2} + 37x - 20 + 8x^{2} - 32x = 5x - 20$

Finding c:

$\frac{5x}{x - 4} = 5$

So c = 5.

Just to verify if the remainder is 0.

$5(x - 4) = 5x - 20$

$5x - 20 - (5x - 20) = 0$

Ok

Then:

$p(x) = (ax^{2} + bx + c)(x - 4)$

$p(x) = (3x^{2} - 8x + 5)(x - 4)$

7 0

3 years ago

( simplify the following polynomials. All final answers must be in standard form )

Mila [183]

$\frac{6x^3-15x^2-3x}{3x} = 2 x^4 - 5 x^3 -x^2$
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3 0

4 years ago

Need help solving this limit problem.

evablogger [386]

Answer:

c= -1.8

Step-by-step explanation:

A function is continuous at a point x if the limits at x from above and below coincide (are equal). Hence,

Limit from above: $\lim_{x\rightarrow\-1^+}f(x)=4c(-1)=6 = -4c-6$

Limit from below: $\lim_{x\rightarrow\-1^-}f(x)=c(-1)^2+3=c+3$

And so when these limits are equal we obtain

$c+3=-4c-6\\5c=-9\\c=-1.8$

3 0

4 years ago