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3 years ago

Find mQ, mR, and mS please

Mathematics

1 answer:

alekssr [168]3 years ago

7 0

It’s a is a freebe !!!!

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The degree of the product is the ____ of the degrees of the factors

Leno4ka [110]

Answer:

Highest

Step-by-step explanation:

This is true because the largest degree is the one that influences the graph and ultimately the end behavior the most. The question you have is simply the a definition-property of degrees.

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3 years ago

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What is the perimeter of rectangle ABCD on the coordinate plane? coordinate grid

Aleksandr-060686 [28]

Just count the number of squares:
(9+5)*2=14*2=28

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3 years ago

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2 x 10 in standard form

lilavasa [31]

Answer:

y = 3 suh slid este ob cine este a, b și g au nevoie de acel kinfo pentru a termina

Step-by-step explanation:

bah de fort

3 0

3 years ago

Pls someone help me find the answers

Alecsey [184]

Answer:

x = 145°

Step-by-step explanation:

We know that y = 98°, so we can find ∠DBC:

180°-y = ∠DBC

180°-98° = 82°

We also know that z = 63°, therefore we can apply <u>exterior angle of triangles</u> to find x.

∠DBC + z = x

82°+63° = x

x = 145°

7 0

3 years ago

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

<u>Step-by-step explanation:</u>

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$

Now, the denominators are same, you can subtract it.

= $\frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

= $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

Thus the simplified solution is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

4 0

3 years ago