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Licemer1 [7]
3 years ago
10

What is slope and y-intercept of the equation on the graph?

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0
The y intercept is the point where the line crosses the y (vertical) axis. So the y intercept is the point (0,3). The slope is 3/2
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Allushta [10]
6 + 8 = 14

1/5 + 4/5 = 5/5 or 1

14 + 1 = 15
4 0
3 years ago
Read 2 more answers
I NEED HELP PLZ I WILL GIVE THANKS AND BRAINLIST
Molodets [167]

Answer:

No

Step-by-step explanation:

Because 50+30 is not 180

5 0
3 years ago
Q - 23 when q = 71 this is evaluating expressions
-BARSIC- [3]
71 - 23 = 48 is the answer
3 0
2 years ago
Can you help me to solve this
stealth61 [152]

Answer:

\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

<u>Triangle 1 (triangle on left):</u>

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is x:x:x\sqrt{2}, where x\sqrt{2} is the hypotenuse of the triangle. Since one of the legs is marked as 30, the hypotenuse must be \boxed{30\sqrt{2}}

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}\\

<u>Triangle 2 (triangle on right):</u>

We can use basic trig for right triangles to set up the following equations:

\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18},

\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

a^2+b^2=c^2, where c is the hypotenuse of the triangle and a and b are two legs of the triangle.

Verify 34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark

3 0
2 years ago
X-2/3= 6 1/3 what is the answer??????????
vichka [17]

For this case we must find the solution of the following equation:

x- \frac {2} {3} = 6 \frac {1} {3}

Rewriting the mixed number in the form of improper fraction we have:

x- \frac {2} {3} = \frac {3 * 6 + 1} {3}\\x- \frac {2} {3} = \frac {19} {3}

Add \frac {2} {3} to both sides of the equation:

x = \frac {19} {3} + \frac {2} {3}

We add the fractions with the same denominator:

x = \frac {19 + 2} {3}\\x = \frac {21} {3}\\x = 7

So, the solution is x = 7

Answer:

x = 7

4 0
2 years ago
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