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Vilka [71]
3 years ago
5

A box has dimensions x+3 and x+7 , find the area of the box

Mathematics
1 answer:
Natali [406]3 years ago
7 0
Multiply (x+3)(x+7)
Foil x^2+10x+21
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Choose the best answer that represents the property used to rewrite the expression.
Andre45 [30]

Answer:

The Quotient Property.

Step-by-step explanation:

Since all three logarithms have the same base (base-5), and you are subtracting 6 and 3, to solve this all you need to do is 6 / 3 because of the Quotient Property.

You aren't multiplying anything, so you wouldn't use the Product Property.

You are not messing around with powers, so you wouldn't use the Power Property.

And you aren't using addition or multiplication, so you wouldn't use the Commutative Property.

Hope this helps!

5 0
3 years ago
What expression can represent 4 more than the product of 6 and 2 1/2
mrs_skeptik [129]

Answer:

6 x 2 1/2 + 4 ??? I hope this is correct :)

Step-by-step explanation:

If its not.. I'm so sorry:(

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7 0
3 years ago
Please answer this question now in two minutes
erastovalidia [21]

Answer: D

154 miles.

Step-by-step explanation:

Half way is 440/2 = 220. Subtract the 66 miles already traveled, and they have 154 miles to go.

7 0
3 years ago
Read 2 more answers
Mr. Ahmed has 31 students in his class. There are 14 boys and 17 girls.
densk [106]
Since there are 14 boys and the whole class has 31 students
We get, the ratio of the part-to-whole relationship for boys is 14/31.

And since there are 17 girls
We get, the ratio of the part-to-whole relationship for girls is 17/31.
7 0
1 year ago
Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet
zmey [24]

Answer:

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

hypotenuse^2=perpendicular^2+base^2

hypotenuse^2=(5)^2+(12)^2

hypotenuse^2=25+144

hypotenuse^2=169

Taking square root on both sides.

hypotenuse=13

In a right angled triangle

\sin \theta = \dfrac{opposite}{hypotenuse}

\sin \theta = \dfrac{5}{13}

\sec \theta = \dfrac{hypotenuse}{adjacent}

\sec \theta = \dfrac{13}{12}

In second quadrant only sine and cosecant are positive.

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

6 0
3 years ago
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