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3 years ago

Molly started her piano lesson at 3:45 pm the lesson lasted 20 minutes.what time did the piano lesson end?

Mathematics

1 answer:

vova2212 [387]3 years ago

4 0

4:05 pm very simple just add 20 minutes to the time

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1. Solve. 6x + 10 = 22 2. Solve. 6x - 4 = 4x What is the answers

Diano4ka-milaya [45]

1) x= 2
6x2+10=22
2) x = 2
6x2-4=4x2

3 0

3 years ago

Question 1 (1 point)

olga55 [171]

Answer:

1) $2y^{2}+4y-9$

2) $18x^{5}y^{4}z$

3) $6x^{5}-12x^{4}+9x^{3}$

4) 40

5) $x^{3}-7x^{2}+3x+36$

Step-by-step explanation:

1) Distribute the negative sign that is outside the parentheses and then you must add like terms, as following:

$(y^{2}-3y-5)-(-y^{2}-7y+4)=y^{2}-3y-5+y^{2}+7y-4=2y^{2}+4y-9$

2) According to the Product property of exponents, when you multiply powers with the same base, you must add the exponents. Then:

$(6xy^{3}z)(3x^{2}yx^{2})=18x^{5}y^{4}z$

3) Apply the Distributive property and the Product property of exponents. Then, you obtain:

$-3x^{3}(-2x^{2}+4x-3)=6x^{5}-12x^{4}+9x^{3}$

4) $(4a+5)^{2}$ is a square of a sum, then, by definition you have:

$(a+b)^{2}=a^{2}+2ab+b^{2}$

Then:

$(4a+5)^{2}=(4a)^{2}+2(4a)(5)+5^{2}=16a^{2}+40a+25$

The coefficient of the second term is the number in front of the variable a. Then, the answer is: 40

5) Apply the Distributive property and the Product property of exponents, then, oyou must add the like terms:

$(x-4)(x^{2}-3x-9)=x^{3}-3x^{2}-9x-4x^{2}+12x+36=x^{3}-7x^{2}+3x+36$

6 0

3 years ago

The new total is $60,but the sales tax is 5% of the bill. How much is the sales tax?

Serhud [2]

3. .5*60 because .5 represents the 5% and "of" stands for multiplication

7 0

3 years ago

Read 2 more answers

Anyone have the answer for this

sergiy2304 [10]

$\huge\bold{Given:}$

Length of the base = 16 km.

Length of the hypotenuse = 34 km. $\huge\bold{To\:find:}$

✎ The length of the missing leg '' $a$ ".

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

The length of the missing leg "a" is $\boxed{30\:km}$ .

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

Using Pythagoras theorem, we have

$({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ ⇢ {a}^{2} + ({16 \: km})^{2} = ({34 \: km})^{2} \\ ⇢ {a}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇢ {a}^{2} = 1156 \: {km}^{2} - 256 \: {km}^{2} \\ ⇢ {a}^{2} = 900 \: {km}^{2} \\ ⇢a \: = \sqrt{900 \: {km}^{2} } \\ ⇢a = \sqrt{30 \times 30 \: {km}^{2} } \\ ⇢a = 30 \: km$

$\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}$

$\huge\bold{To\:verify :}$

$( {30 \: km})^{2} + ({16 \: km})^{2} =( {34 \: km})^{2} \\ ⇝900 \: {km}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\⇝1156 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇝L.H.S.=R. H. S$