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2 years ago

(2t^2+13t+15)\(t+6) long division

Mathematics

1 answer:

aleksklad [387]2 years ago

6 0

Answer:

I think this is the correct solution

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Please Help! 5.1 Algebra

sasho [114]

Answer:

Step-by-step explanation:

Since x is in an absolute value, we know that no matter what value we put in for x, the result will always be positive. B fits this description.

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2 years ago

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Can someone answer this question please please help me I really need it if it’s correct I will mark you brainliest .

MrRissso [65]

Formula for the area of a face of a cube: A = length x width

Let's find the area of one face.

A = 5 x 5

A = 25

Now, let's multiply the area of one face by 6 because a cube has 6 faces.

25 x 6 = 150

A = 150 square centimeters

Hope this helps! :)

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3 years ago

Solve for x

Olenka [21]

X is equal to -4. Hope I helped. If you want an explanation too I would be glad to provide that .

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3 years ago

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Are women better at multitasking than men? Researchers conducted a multitasking experiment in which 120 men and 120 women played

stich3 [128]

Answer:

Yes, women better at multitasking than men.

Step-by-step explanation:

Given

$n_1=120$ --- Men

$n_2=120$ --- Women

$\bar x_2 > \bar x_1$ --- Women with greater average speed

Required

Does women better at multitasking?

When making comparison between two variables that are subjected to the same conditions;

If both variables have different mean values; the variable with the greater mean value performs better.

So, in this case:

$\bar x_2 > \bar x_1$ means that women perform better than men, at multitasking.

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2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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3 years ago