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3 years ago

In right △ABC with right angle B, m∠A=(3x−8)° and m∠C=(x−2)°.

Mathematics

2 answers:

Crank3 years ago

8 0

Answer:

Option 2nd is correct

67°

Step-by-step explanation:

Given that:

In right △ABC with right angle B.

$m \angle B = 90^{\circ}$ , m∠A=(3x−8)° and m∠C=(x−2)°.

We know that the sum of all the measures of a triangle is 180 degree

In triangle ABC

⇒ $m\angle A+ m\angle B+ m\angle C = 180^{\circ}$

Substitute the values we have;

$3x-8+90^{\circ}+x-2 = 180^{\circ}$

Subtract 90 degree from both sides we have;

$3x-8+x-2 = 90^{\circ}$

Combine like terms;

$4x-10= 90^{\circ}$

Add 10 to both sides we have;

$4x=100^{\circ}$

Divide both sides by 4 we have;

$x=25^{\circ}$

then;

m∠A=(3(25)−8)° = (75-8)° = 67°

Therefore, the measure of angle A is, 67°

Lubov Fominskaja [6]3 years ago

5 0

Add both and set it equal to 90

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Katena32 [7]

Hello!

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3 years ago

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The numbers on two consecutively numbered gym lockers have a sum of 141. What are the locker numbers?

Tamiku [17]

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3 years ago

What is the value for y

egoroff_w [7]

Answer:

y = 107°

Step-by-step explanation:

As you can see from the markings on the triangle, this is an isosceles triangle (two sides are congruent).

An isosceles triangle has the base angles congruent too.

So at this point,

34 = x - 5

Solve for x,

34 = x - 5

x = 34 + 5

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Now, the total angle measure of triangle is 180°.

There are two angles ready to solve for y.

34 + 39 + y = 180

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I would like the answer to 1/4 minus 1/3 is it 12

Mrac [35]

Answer:

-1/12

Step-by-step explanation:

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2 years ago

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a

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$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

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This means that $f$ has a non-removable discontinuity at $x=3$ . Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$ -intercept at (1, 0)

This means $f(1)=0$ .

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$ . The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.