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3 years ago

Which dimensions can create only one unique triangle?

Mathematics

1 answer:

Anon25 [30]3 years ago

5 0

Answer:

C. three sides measuring 5 in, 12 in, and 14 in.

Step-by-step explanation:

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Bradley is returning home from a place that is 2 kilometers away. The function y = 2,000 − 90x represents Bradley's distance fro

12345 [234]

Answer:

The function is linear

The function changes at a constant rate

Step-by-step explanation:

If this function were to be graphed, a line would be seen. It is not an increasing function because he is getting closer to home so the distance from his house is decreasing

6 0

3 years ago

Balloon Payment (Math)

fiasKO [112]

Answer:

Step-by-step explanation:

The loan duration is for 8 years, a total of 12·8 = 96 monthly payments. The 96th payment is the balloon payment.

7 0

3 years ago

33 1/3% of 75<br><br>plz help

Paraphin [41]

The answer is 25

x 33 1/3% = 75

75 / 33 1/3

0.3333

(75)(0.333) = 25

24.9999999 rounds to 25

Hope this helps, have a BLESSED day! :-)

6 0

3 years ago

Solve the equation -25+4y=35

borishaifa [10]

-25+4y=35
*Subtract 35*
-25+4y-35=0
*add 25 (to cancel out)*
4y-35=25
*add 35 (to cancel out)*
4y=60
*Divide by 4 to get y on its own*
y=15
Hope this helps :)

6 0

3 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago