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What is 10 1/2 ÷ 3 1/5

Igoryamba

It’s 3.2812 then round u get 3.3012

4 0

4 years ago

MP bisects ∠BMS. If m∠BMP = 2x + 9 and m∠BMS = 7x – 3, find the value of m∠PMS

dimulka [17.4K]

Answer:

Step-by-step explanation:

If MP bisects ∠BMS, then the line MP divides <BMS equally;

The adition postulate if therefore true;

<BMP +< PMS = <BMS and <BMP = < PMS

The equation becomes;

<BMP +< BMP= <BMS

2 <BMP = <BMS

2(2x+9) = 7x - 3

4x+18 = 7x-3

collect like terms

4x-7x = -3-18

-3x = -21

x = 21/3

x = 7

Since <BMP = < PMS 2x+9

< PMS = 2(7)+9

< PMS = 14+9

< PMS = 23

Hence the value of < PMS is 23

3 0

3 years ago

Point Q is the centroid of △ABC. QF = _____ A. 10 B. 5 C. 15 D. 9

Keith_Richards [23]

Answer: QF = 5

Step-by-step explanation:

We know that if a triangle has a centroid, the ratio of the longer segment to shorter segment is 2:1. We can set up a proportion $\frac{2}{1} = \frac{10}{x}$ , where x is the length of QF. By cross multiplying and dividing, you get x = 5 or QF = 5.

6 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

a)

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

a)

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$