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brilliants [131]
3 years ago
10

I can't seem to answer number 23? How do you solve it?

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
3 0
For every 10 degrees Celsius, degrees Fahrenheit goes up 18, so
10 degrees C is 50
20 degrees C is 68
30 degrees C is 86
40 degrees C is 104
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Points P, Q, and S are collinear . If bisects PQR, what is the measure of PQT?
stepladder [879]
719.81 and dont mind the rest of this it said i need 20 words lol and can u do my question plz ?
5 0
2 years ago
The outer and inner triangles are both equilateral and the circle touches all three sides of the outer triangle. If the area of
Brrunno [24]

Answer:

The area of outer Δ = 40 cm²

Step-by-step explanation:

∵ Area of the small triangle = 10 cm²

If we join the center of the circle with the 3 vertices of the inner Δ

These 3 segments are the radii of the circle

Now the inner triangle has 3 isosceles Δ their sides are r , r and s1 with vertex angle 120° ⇒ (360° ÷ 3 = 120°)

Where r is the radius of the circle and s1 is the side of the inner triangle

<em>By using cosine rule</em>

(s1)² = r² + r² - 2r²cos120 = r² + r² - 2r² (-0.5) = r² + r² + r² = 3r²

∴ s1 = r√3

∵ The radius of the circle ⊥ to the side of the outer Δ because the side of the outer Δ is a tangent to the circle

If we join a vertex of the outer Δ with the center of the circle

We will have a right angle triangle of two legs r and half s2 with angle 60° (120° ÷ 2 )between them ⇒ s2 is the side of outer Δ

∴ tan 60° = 1/2 (s2) ÷ r ⇒ √3 = 1/2 (s2) ÷ r = (s2)/2r

∴ s2 = 2r√3

∴ s2 : s1 = 2r√3 ÷ r√3 = 2 : 1

∴ The side of the outer Δ is double the side of the inner Δ

<em>By using similarity ratio</em>

A2/A1 = (s2/s1)² ⇒ A2 and A1 are the areas of outer and inner triangles

∴ A2 : A1 = (2/1)² = 4/1  

∴ A2 = 4 A1

∴ A2 = 4 × 10 = 40 cm²

5 0
3 years ago
How do I graph this slope?
maria [59]
You can use the equation rise over Run you go up 4 and over 6 and you go down 2 and over 8 (im not sure if this helps)
5 0
2 years ago
If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?
Rufina [12.5K]
x is in quadrant I, so 0, which means 0, so \dfrac x2 belongs to the same quadrant.

Now,

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

Since \sin x=\dfrac5{13}, it follows that

\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}

Since x belongs to the first quadrant, you take the positive root (\cos x>0 for x in quadrant I). Then

\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}

\tan x is also positive for x in quadrant I, so you take the positive root again. You're left with

\tan\dfrac x2=\dfrac15
4 0
3 years ago
Instructions are in the picture
Phantasy [73]

Answer:

123123 3213123 12312 dasdsd aw dasd sda asdasd

Step-by-step explanation:

3 0
3 years ago
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