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3 years ago

Identify the unlike terms in the expression: 3a + 11b + 30 + 4a + 3b

Mathematics

2 answers:

MAVERICK [17]3 years ago

6 0

Answer: 7a+14b+30

How to: Combine like terms

Have a great day and stay safe ! sorry if im wrong

erastovalidia [21]3 years ago

3 0

30 is the unlike term cause it doesn’t have a variable match like the others i don’t know if I did it wrong I’m sorry

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Work out the area of this shape. ? cm2 4 cm 9 cm 6 cm 12 cm Help!!!

mars1129 [50]

Answer:

84 cm^2

Step-by-step explanation:

Area of triangle + area of rectangle

Area of triangle = 1/2*l*b

= 1/2*8*3= 12 cm^2

Area of rectangle = l*b

= 12*6 = 72 cm^2

Total area= 72+12

= 84 cm^2

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3 years ago

Is my answer correct

Murljashka [212]

Yes your answer should be c;

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3 years ago

Which terms complete the factorization of x2 27x 162 represented by the model? 27, 9x, 18x 9, 9x, 18x 27, 9x, 27x 9, 9x, 27x

olga nikolaevna [1]

Answer: 9, 9x, 18x

Step-by-step explanation:

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2 years ago

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Sedaia [141]

Answer: I think it’s 225
Explanation: 1,500 • .15= 225

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3 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

6 0

3 years ago

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