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4 years ago

13

it takes jordan 15 minutes to do 12 math problems. how many problems will he finish if he works for 1.5 hours? what is jordan's

average rate of math problems completed per hour?

2 answers:

shtirl [24]4 years ago

8 0

6 I think but I am not great at math

Nitella [24]4 years ago

7 0

He will finish 72 problems and I think the average rate is 48 or 4. not sure about the average

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make a hypothesis about the sum of the interior angles of any trangle. Explain why your hypothesis works for all triangles

lakkis [162]

Answer:

The sum of the three angles of any triangle will always be 180 degrees because when all three angles are put together, they form a straight line.

Step-by-step explanation:

I did it on Edgenuity

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3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

Whats the step by step process to solve 3(×-9)=30

Sergio [31]

<span>step 1 :</span> 3 • (x - 9) - 30 = 0 <span>Step 2 :</span><span>Step 3 :</span>Pulling out like terms :

<span> 3.1 </span> Pull out like factors :

3x - 57 = 3 • (x - 19)

<span>Equation at the end of step 3 :</span> 3 • (x - 19) = 0 <span>Step 4 :</span>Equations which are never true :

<span> 4.1 </span> Solve : 3 = 0

<span>This equation has no solution.
</span>A a non-zero constant never equals zero.

Solving a Single Variable Equation :

<span> 4.2 </span>     Solve  :    x-19 = 0<span>

</span>Add  19 to both sides of the equation :<span>
</span>                     x = 19

One solution was found : <span> x = 19</span>

7 0

3 years ago

Read 2 more answers

How many 5 digit numbers are even if repetition is not allowed?

shtirl [24]

23308 is the answer of how many is even with no repitition

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4 years ago

In ΔOPQ, o = 9.2 cm, p = 2.4 cm and ∠Q=37°. Find the length of q, to the nearest 10th of a centimeter.

storchak [24]

The length of q, to the nearest 10th of a centimeter is 7.6 cm.

Given in question,

In ΔOPQ,

o = 9.2 cm

p = 2.4 cm

∠Q = 37°

Cosine formula ⇒ cos θ = $\frac{o^{2}+p^{2}-q^{2} }{2op}$

Putting the values in equation,

cos 37 = $\frac{(9.2)^{2}+(2.4)^{2}-q^{2} }{2*9.2*2.4}$

0.799 = $\frac{84.64 + 5.76-q^{2} }{44.16}$

0.799*44.16 = 90.4 - $q^{2}$

32.28 = 90.4 - $q^{2}$

$q^{2}$ = 90.4 - 32.28

$q^{2}$ = 58.12

$q = \sqrt{58.12}$

$q = 7.63$

q = 7.6 cm (to nearest 10th)

Hence, length of q is 7.6 cm.

Learn more about length on:

brainly.com/question/8552546

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3 0

2 years ago