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3 years ago

Please tell me if I'm right or wrong! No work needed! Brainliest will be given!

Mathematics

2 answers:

Alex777 [14]3 years ago

4 0

Answer:

The first one is correct

The second one is also correct

The third is also correct

Congrats!

N76 [4]3 years ago

3 0

Answer:

1) $\boxed{Option \ B}$

2) $\boxed{Option \ B}$

3) $\boxed{Option \ B}$

You're totally correct, Man! :)

Step-by-step explanation:

Question 1:

$(6b^2-4b+3)-(9b^2-3b+6)\\Resolving \ the\ brackets\\6b^2-4b+3-9b^2+3b-6\\Combining \ like \ terms\\6b^2-9b^2-4b+3b+3-6\\-3b^3-b-3$

Question 2:

$(b+6)(b-3)\\Using \ FOIL\\b^2-3b+6b-18\\b^2+3b-18$

Question 3:

$(4x-3)(6x-1)\\Using \ FOIL\\24x^2-4x-18x+3\\24x^2-22x+3$

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At an ice cream store, a family ordered 4 banana splits and 5 hot fudge sundaes, paying a total of $27 for their order. The next

mash [69]

Answer:

both items are $3 each

Step-by-step explanation:

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3 years ago

What is the slope of a line that passes through (6,4) and (-1, 3)

pshichka [43]

Answer:

C) 1/7

Step-by-step explanation:

so idk what the formula is called but it looks like this:

y2-y1/x2-x1 and if you plug in the numbers then you'll get the answer

3 - 4/-1 - 6 = -1/-7 = 1/7

you can also switch the numbers your answer will always be the same

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so the answer is c

4 0

4 years ago

What is the value of y?

Molodets [167]

B, because
3/15=x/75
75/15=5
3*5=15
15 is your answer.

4 0

3 years ago

The sum of (2a + ab + 2b) + (4a2 - 3ab + 9)

Sati [7]

You can only add/subtract like terms so:

2a+ab+2b+4a^2-3ab+9

2a-2ab+2b+4a^2+9 (the only like terms were ab and -3ab)

4 0

4 years ago

Find the area of the triangle belowunits given are cm, cm^2, cm^3

zloy xaker [14]

Answer:

$A_T=150cm^2$

Explanation: We need to find the area of the triangle, We know that the area of any right triangle is:

$A_T=\frac{1}{2}(b\times h)$

Therefore in the information given we have the following:

$\begin{gathered} h=20\operatorname{cm} \\ b=15\operatorname{cm} \end{gathered}$

Therefore the Area of this triangle is as follows:

$\begin{gathered} A_T=\frac{1}{2}(b\times h) \\ \therefore\rightarrow \\ A_T=\frac{1}{2}(20cm\times15cm)=\frac{1}{2}(20\times15)cm^2 \\ A_T=\frac{1}{2}(300)cm^2=150cm^2 \\ \therefore\rightarrow \\ A_T=150cm^2 \end{gathered}$

Note! Units are centimeter-squared

4 0

1 year ago