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2 years ago

Helppppppp:))))))).......

Mathematics

1 answer:

-BARSIC- [3]2 years ago

5 0

Answer:

y=6 is the correct answer

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HELP DUE IN 15 MINS!<br><br> x=??

ad-work [718]

Answer:

Step-by-step explanation:

Remark

The angle formed by the intersection of a tangent to a circle and an intersecting secant is 1/2 (Arc PS - arc PR)

PS = 208

PR = 74

Solution

x = 1/2 (208 - 74)

x = 1/2 (134)

x = 67

4 0

2 years ago

What fraction is equivalent to 36/60 and the sum of the numerator and denominator is 24

almond37 [142]

Answer:

The fraction is $\frac{9}{15}$

Step-by-step explanation:

$\frac{36}{60}$ = $\frac{x}{y}$

Where x and y are the numerator and denominator of our new fraction respectively.

Also given is that x + y = 24

So $\frac{x}{y}$ = 36 ÷ 60 = 0.6

x = 0.6y

Substituting x in x + y = 24 gives;

0.6y + y = 24 , 1.6y = 24

y = 24 ÷ 1.6 = 15

x = 24 - 15 = 9

So our fraction is $\frac{9}{15}$

5 0

3 years ago

Help me find the surface area of the candle

ExtremeBDS [4]

Answer:

Easy!

Step-by-step explanation:

1) To find the surface area of a regular triangular pyramid, we use the formula SA = A + (3/2)bh, where A = the area of the pyramid's base, b = the base of one of the faces, and h = height of one of the faces.

2)Multiply the side length of the base by the slant height and divide by two. Then, multiply by 4. This will give you the lateral surface area of the pyramid. Add the base surface area and the lateral surface area.

4 0

3 years ago

PLEASE HELP!! asap!!

spin [16.1K]

Answer:

The third one from left to right

Step-by-step explanation:

Just check that when x = 11

y= $-\sqrt[3]{11-3}+4=-2+4=2$

So the graph passes through the point (11,2)

7 0

3 years ago

Read 2 more answers

What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?

xxMikexx [17]

We have been given an equation of hyperbola $\frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1$ . We are asked to find the center of hyperbola.

We know that standard equation of a vertical hyperbola is in form $\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

$\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1$

Now we can see that value of k is $-3$ . Therefore, the vertex of given hyperbola will be at point $(6,-3)$ and option D is the correct choice.

6 0

3 years ago