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Art [367]
3 years ago
14

Which expression is equivalent to the following expression: 15 + 6x A. 3(5 +2x) B. 3(5 + 6x) C. 21x D. 15(6x)​

Mathematics
1 answer:
Makovka662 [10]3 years ago
7 0
It’s on photomath ifyou have it
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Help me please, I’ve been stuck on this. :(
maxonik [38]

Answer: 0.98 cents

Step-by-step explanation:

First you want to get the amount of apples to one pound.

3.5 to 1

3.5 divided by 3.5 equals to one.

Now you need to o the same thing to the cost

3.43 divided by 3.5

= 0.98

since we are talking money it would be 0.98 cents

4 0
2 years ago
I REALLY NEED HELP! What is the length of the meteor with the most common recorded value? Select one: a. four eighths inch b. 1
Triss [41]

Answer:

Step-by-step explanation:

most common recorded value= 2inches optiond

hope it helps

5 0
3 years ago
Read 2 more answers
Please i beg u help me, this will mean a lot!
slega [8]

Answer:

17

Step-by-step explanation:

Mode is the number that appears the most

3 0
2 years ago
Someone please help me with this
Nadusha1986 [10]
Since 2 sides of the triangle are the same, it's an isosceles triangle.
That would also mean that the two angles are the same; 70 and 70.

A triangle has 180 degrees total.

70 + 70 + x = 180
x = 40
8 0
3 years ago
Read 2 more answers
Given: △ABC, AB=5 square root 2 <br><br>m∠A=45°, m∠C=30°<br><br>Find: BC and AC
hichkok12 [17]

Answer:

Part a) BC=10\ units

Part b) AC=13.66\ units

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

\frac{AB}{sin(C)}=\frac{BC}{sin(A)}

we have

AB=5\sqrt{2}\ units

A=45^o

C=30^o

substitute

\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}

solve for BC

BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))

BC=10\ units

step 2

Find the measure of angle B

we know that

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

so

m\angle A+m\angle B+m\angle C=180^o

substitute the given values

45^o+m\angle B+30^o=180^o

75^o+m\angle B=180^o

m\angle B=180^o-75^o

m\angle B=105^o

step 3

Find the length side AC

Applying the law of sines

we know that

\frac{AB}{sin(C)}=\frac{AC}{sin(B)}

we have

AB=5\sqrt{2}\ units

A=45^o

B=105^o

substitute

\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}

solve for AC

AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))

AC=13.66\ units

4 0
2 years ago
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