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3 years ago

When multiplied together, which two polynomials equal ab3+2a2b2? Select the two correct answers.

Mathematics

2 answers:

fomenos3 years ago

5 0

The answer is 2) ab2 that would be ur answer

g100num [7]3 years ago

5 0

Answer:

The Answer to the question would be option 1.) b+2a, and option 2.) ab^2

Step-by-step explanation:

Here is the equations and all that that would be used:

ab2(b+2a)

=(ab2)(b+2a)

=(ab2)(b)+(ab2)(2a)

=ab3+2a2b2

=2a2b2+ab3

With this, we find that multiplying:

ab^2 * (b + 2a), we will arrive at our answer.

Really hope this helps!

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Doug’s go-kart can go 50 m in 20s. What is its average speed?

Olegator [25]

Answer:

2.5m

Step-by-step explanation:

50m divided by 20 = 2.5m

4 0

3 years ago

there are 300 student and teachers at a sports event.Grade k to 2 is students: 137 Grade 3 to 5 students: 145 Teachers: ? what i

AVprozaik [17]

Answer:

Percentage of teachers = 6%

Step-by-step explanation:

Given;

Number of grade k-2 students = 137

Number of grade 3-5 students = 145

Total = 300

Number of teachers = 300 - (137+145) = 18

Percentage of teachers = (number of teachers÷total)× 100%

%T = (18/300) × 100

%T = 6%

8 0

3 years ago

Quadrilateral ABCD is a square. if m

olganol [36]

The answer is B. 18

Explanation:

We know that a square has 4 right angles. A right angle is 90°.

5x° = 90°
5(18)° =90°
90°=90°

Hope this helps some

4 0

3 years ago

20 POINTS: solve for x.

lara [203]

Answer:x=7

Step-by-step explanation:

Here AD=6

AB=6 (tangents drawn from external point to a circle are equal)

Then BC=13-6=7

Now, CE=BC=7(tangents drawn from external point to a circle are equal)

EF=14-7=7

EF=FG=x=7(tangents drawn from external point to a circle are equal)

3 0

3 years ago

Plz show work and solve quick please need help!!!!!

Simora [160]

Surface area is just the area of all these 4 triangles plus the rectangle.

First we can find the area of the rectangle.

$\sf A=lw$

Half of the length is 28 cm, so the full length must be 28 * 2 = 56 cm.

$\sf A=(56)(27)$

$\sf A=1512cm^2$

The base for the left and right triangles are 27. The heights would be the net length minus half the length of the rectangle:

$\sf 82.8-28=54.8\? cm$

Calculate the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(27)(54.8)$

$\sf A=739.8\?cm^2$

We have two of these triangles.

$\sf 739.8\times 2=1479.6\? cm^2$

Now do the other two pair of triangles. The bases for them are 28 + 28 = 56 cm. The heights would be the net width minus the width of the rectangle:

$\sf 81.2-27=54.2\?cm$

Now find the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(56)(54.2)$

$\sf A=1517.6\? cm^2$

We have two of these triangles.

$\sf 1517.6\times 2=3035.2\?cm^2$

Add all the areas together:

$\sf 1512+1479.6+3035.2=\boxed{\sf 6026.80\? cm^2}$

7 0

3 years ago