Find the least common denominator of these two fractions. LOTO and 1/ 5 The least common denominator is [?]

Mathematics

1 answer:

Oksana_A [137]2 years ago

3 0

Answer:

6yx

Step-by-step explanation:

This number will be your least common denominator (LCD). Divide the LCD by the original denominator. To determine the multiple needed to make the denominators equal, divide the LCD you determined by the original denominator. Multiply the numerator and the denominator of each fraction by this number.

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5-1-9/3-9 What do I do?

Contact [7]

Rewrite it to simplify using order of operations.

P E M D A S
() ^2 × ÷ + -

So nw it's
5-1-(9÷3)-9

no now it's
(5-1)-(3-9)
4-(-6) I wrote the negative in parentheses to show its not an additional subtraction sign. also if you look it's sort of a plus sign. so when you subtraction (-6) from 4 it's actually 4+6

4-(-6)=10.

that's what I got. hope this helps.

7 0

3 years ago

Read 2 more answers

The area of sector A08 is 48x and m<AOB = 270º. Find the radius of circle O.

Softa [21]

Answer:

Step-by-step explanation:

The first one gives us everything we need except the radius, which is easy enough to solve for if you're careful with your algebra. The area of a sector of a circle is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ is the measure of the central angle of the circle. For us, that fills in as follows:

$48\pi=\frac{270}{360}*\pi r^2$ and manipulate it as follows:

$r^2=\frac{(360)(48\pi)}{270\pi}$ the π's cancel out, leaving us with simple multiplication and division to get

r = 8. Now for the next one, which is a bit more involved.

In order to find the area of the shaded part, we need to find the area of the right triangle there and subtract it from the area of the sector of the circle. First the area of the sector, which is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ again is the measure of the central angle of the circle, 90°:

$A_s=\frac{90}{360}(3.1415)(10)^2$ which simplifies a bit to

$A_s=\frac{1}{4}(3.1415)(100)$ , giving us an area of

$A_s=78.5375m^2$ . Now onto the area of the triangle.

Since this triangle is inscribed in the circle and the circle's radius is 10, tha also gives us both the height and the base measures of the triangle. The area then is:

$A_t=\frac{1}{2}(10)(10)$ which is