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lina2011 [118]
3 years ago
9

The measures of the angles of △ABC are given by the expressions in the table. Angle Measure A ​ (4x − 13)° ​ B 15° C (x + 18)∘ W

hat is the measure of ∠A and ∠C ?
Mathematics
1 answer:
liubo4ka [24]3 years ago
5 0
By Angle sum property,
Sum of all angles of a triangle equal to 180°.

As per the property,
A+B+C = 180°
4x-13 + 15 + x+18 = 180
5x+20 = 180
5x = 160
x = 32

Measure of Angle A =>
4x-13 = 4*32-13 = 128-13 = 115°

Measure of Angle C =>
x+18 = 32+18 = 50°

!! Hope It Helps !!
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Lines sand tare perpendicular. If the slope of line sis 5, what is the slope of<br> line t?
Sveta_85 [38]

Answer:

  -1/5

Step-by-step explanation:

Perpendicular lines have slopes that are the opposite reciprocal of one another.

The slope of line t is the opposite reciprocal of the slope of line s:

  slope of t = -1/(slope of s)

  slope of t = -1/5

6 0
3 years ago
What is the area of a circle with a diameter of 12.6 in.?
Zepler [3.9K]

Hey there! I'm happy to help!

To find the area of a circle, you square the radius and then multiply by pi (3.14 in our case).

The radius is half of the diameter.

12.6/2=6.3

We square this.

6.3²=39.69

We multiply by 3.14

39.69×3.14=124.6266

We round to the nearest hundredth, giving us an area of 124.63 in².

Now you can find the area of a circle! Have a wonderful day! :D

5 0
3 years ago
Find the volume of the triangular prism in the picture shown. Round your answer to the
Olenka [21]

Given:

A figure of a triangular prism.

Height of triangular base = 9\dfrac{2}{3} yd

Base of triangular base = 4 yd

Height of prism = 4 yd.

To find:

The volume of the triangular prism.

Solution:

Volume of a triangular prism is:

V=Bh              ...(i)

Where, B is the area of triangular base and h is the height of the prism.

Area of triangular base is:

B=\dfrac{1}{2}\times base\times height

B=\dfrac{1}{2}\times 4\times 9\dfrac{2}{3}

B=2\times \dfrac{29}{3}

B=\dfrac{58}{3}

Putting B=\dfrac{58}{3} and h=4 in (i), we get

V=\dfrac{58}{3}\times 4

V=\dfrac{232}{3}

V=77.33...

V\approx 77.3

The volume of the triangular prism is 77.3 cubic yards. Therefore, the correct option is 2.

4 0
2 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0
3 years ago
Solve: 9x2 - 30x + 20 = 0
Trava [24]
Actually I don't know the answer but what I thing is none of the above.
4 0
3 years ago
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