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Naily [24]
2 years ago
10

WOULD YOU RATHER...

Mathematics
2 answers:
asambeis [7]2 years ago
8 0

Answer: first one

Step-by-step explanation:

8090 [49]2 years ago
7 0

Answer:

0.015 km/s

Step-by-step explanation:

40 km/hr

0.015 km/s

0.015 ÷ (1/3600)

0.015 × 3600

= 54 km/hr

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Find each missing measure
Trava [24]
Could you try sending the picture again? I don't see anything. Sorry.
8 0
3 years ago
Solve x+y=7 and 3x-2y=11 using elimination
BartSMP [9]
X=5 and y =2.
Here's how you do it.
Multiply the first equation by -3, and the second one by 1.
add the equations to eliminate x.
then solve for y by dividing both sides by -5, which gives us 2.
to solve for x subsitute 2 in for y.

8 0
3 years ago
I don’t understand what this means
Colt1911 [192]

Answer:

This means that if you have that amount of those and you do whatever the rest of the question is.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
4x/5 - 3x/7 = 4x/7 + 5x/7 can anyone please help with explanation...​
kirill115 [55]

Answer: x=0

Step-by-step explanation:

Multiply both sides of the equation by 35, the least common multiple of 5,7.

7×4x−5×3x=5×4x+5×5x

Multiply 7 and 4 to get 28.

28x−5×3x=5×4x+5×5x

Multiply −5 and 3 to get −15.

28x−15x=5×4x+5×5x

Combine 28x and −15x to get 13x.

13x=20x+25x

Combine 20x and 25x to get 45x.

13x=45x

Subtract 45x from both sides

13x−45x=0

Combine 13x and −45x to get −32x

−32x=0

Product of two numbers is equal to 0 if at least one of them is 0. Since −32 is not equal to 0, x must be equal to 0.

X=0

7 0
2 years ago
In triangle ΔABC, ∠C is a right angle and CD is the height to
Zina [86]

Answer:

m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles m\angle CDB=m\angle CDA=90°

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha

From ΔCBD,

m\angle CBD is same as m\angle B.

So, m\angle CBD=90-\alpha

m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha

Now, from ΔCAD,

m\angle CAD is same as m\angle A

So, m\angle CAD=\alpha

m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha

Hence, the unknown angles of both the triangles are:

m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha

5 0
2 years ago
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