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3 years ago

Which is a simplified form of the expression 6(2t – 3) + 3(t + 5)?

Mathematics

2 answers:

faust18 [17]3 years ago

8 0

Answer:

it's option D.

Step-by-step explanation:

see attachment in case u need steps

Ostrovityanka [42]3 years ago

6 0

D would be you answer

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Solve for x in the equation x squared + 2 x + 1 = 17.

lbvjy [14]

Answer:

Step-by-step explanation:

x^2 + 2x + 1 = 17

x^2 + 2x = 17 - 1

x^2 + 2x = 16

x^2 + 2x + 1 = 16 + 1

(x + 1)(x + 1) = 17

(x + 1)^2 = 17

x + 1 = sqrt 17

x = -1 (+ - ) sqrt 17

x = -1 + sqrt 17 or x = -1 - sqrt 17 <===

7 0

3 years ago

Mario purchased a computer with a screen that measures 16.0 inches diagonally. The length of the screen is 11 inches. What is th

Vesnalui [34]

Answer:

11.61 in

Step-by-step explanation:

the screen when divided diagonally, forms 2 right triangles

diagonal = hypotenuse = 16

16^2 = 11^2 + w^2

w^2 = 135

w = 11.61

3 0

2 years ago

Read 2 more answers

Read the sentence.

AfilCa [17]

Answer:

I think it's simple sentence

6 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago

So i basically have to put true or false, please give me the right answer:

4vir4ik [10]

Answer:

Yes, No, Yes, No.

Step-by-step explanation:

3 0

3 years ago