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Svetllana [295]
3 years ago
6

Anyone know how to do this?

Mathematics
1 answer:
Reika [66]3 years ago
5 0

Answer:

-2x-y-15

Step-by-step explanation:

5x+3y+4-9x-8+2x-4y-11

you add the x together

5x+2x=7x

7x-9x=-2x

then yourewrite the problem

-2x+3y+4-8-4y-11

then you can solve for the y

3y-4y= -1y but you do not need to put the 1 since -y is the same as -1y

we can rewrite the problem again

-2x-y+4-8-11

4-8=-4

-4-11=-15

answer

-2x-y-15

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S= savings before she bought the dress   s - $55 would be her new balance after she bought the dress.

We could also talk about her savings now (after her purchase).   If we let n=now
n + $55 would be her balance before she bought the dress.
8 0
4 years ago
Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian
777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\B=30^o

Applying the law of sines

Find the measure of angle A

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(A)}=\frac{2}{sin(30^o)}

sin(A)=1

so

A=90^o

Find the measure of angle C

In a right triangle

we know that

B+C=90^o ----> by complementary angles

B=30^o

therefore

C=60^o

Find the length side c

Applying the law of sines

\frac{c}{sin(C)}=\frac{b}{sin(B)}

substitute the given values

\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}

c=2\sqrt{3}\ units

therefore

The dimensions of the triangle are

A=90^o

B=30^o

C=60^o

a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units

<em>Second case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\A=30^o

Applying the law of sines

Find the measure of angle B

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(30^o)}=\frac{2}{sin(B)}

sin(B)=0.25

so

using a calculator

B=14.48^o

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

A+B+C=180^o

A=30^o\\B=14.48^o

therefore

30^o+14.48^o+C=180^o

C=135.52^o

Find the length side c

Applying the law of sines

\frac{c}{sin(C)}=\frac{a}{sin(A)}

substitute the given values

\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}

c=5.61\ units

therefore

The dimensions of the triangle are

A=30^o

B=14.48^o

C=135.52^o

a=4\ units\\b=2\ units\\c=5.61\ units

see the attached figure to better understand the problem

4 0
3 years ago
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The altitude of the kite:
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Answer:

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TIMED PLEASE HELP ASAP! MULTIPLE CHOICE, BRAINLIEST FOR CORRECT ANSWER!
erma4kov [3.2K]
I think it is the third one
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3 years ago
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