A product of two (or more) factor can be zero if and only if at least one of the factors is zero.
In other words, you cannot multiply two non-zero real numbers, and have zero as a result.
So, if we want the product of these two factors to be zero, at least one of them has to be zero.
The first factor is zero if
The second factor is zero if
Answer:
Step-by-step explanation:
Let me know if you want the full explanation. Have a great day! ❤
Mia walked 7km/h so after 1 hour, she is 7 km north of the house of Julia
Samantha walked 11km/h so after 1 hour, she is 11 km west of the house of Julia.
The points where Mia and Samantha are after 1 hour , and the house of Julia form a right triangle with sides 7 and 11 km. The distance between the girls, is the hypotenuse of his triangle.
by the pythagorean theorem:
Answer: 13 km
(Pls give brainliest! :D)
Answer:
<em>B</em>
Step-by-step explanation:
<em>You know that the graph intercepts at the point of (4,-2)</em>
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<em>Where x = 4 and y = -2.</em>
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<em>This means that whenever you plug a value of x in the equation it should give the value of y.</em>
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<em>For example when you plug x= 4 into the equation (B), it should look like this.</em>
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<em>4- y = 6, solving for y gives us the value negative 2.</em>
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<em>This means that it does give us this point, however, as the graph shows us, both lines meet at the point which means that both equation should give us this point.</em>
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<em>If we substitute x = 4 in the second equation of b, we should be able to get 3(4) + 4y = 4. Solving for y gives us -2.</em>
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Hope this helps!
-<em>Yumi</em>
Let's work on the left side first. And remember that
the<u> tangent</u> is the same as <u>sin/cos</u>.
sin(a) cos(a) tan(a)
Substitute for the tangent:
[ sin(a) cos(a) ] [ sin(a)/cos(a) ]
Cancel the cos(a) from the top and bottom, and you're left with
[ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a)</u> ] That's the <u>left side</u>.
Now, work on the right side:
[ 1 - cos(a) ] [ 1 + cos(a) ]
Multiply that all out, using FOIL:
[ 1 + cos(a) - cos(a) - cos²(a) ]
= [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>.
Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.
So, on the <u>right side</u>, you could write [ <u>sin²(a)</u> ] .
Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> .
They look quite similar. In fact, they're identical. And so the identity is proven.
Whew !
