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2 years ago

Enrique earns 10 points for each question that he answers correctly on a geography test.

Mathematics

2 answers:

Lemur [1.5K]2 years ago

6 0

Answer: y =10x

Step-by-step explanation:

Hi, to answer this question we have to write an equation with the information given:

The total number of points (y) will be equal to the number of questions he answers correctly (x) and the points earned from each correct answer (10).

Mathematically speaking:

y =10x

Feel free to ask for more if needed or if you did not understand something.

saul85 [17]2 years ago

4 0

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ The equation would be y = 10x

➶ Hope This Helps You

➶ Good Luck

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Which problem situation matches the equation?

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Habib is making some scones.

sergeinik [125]

Answer:

$180^\circ C$

Step-by-step explanation:

To convert from degree Fahrenheit to Celsius, we use the formula:

$^\circ C=\dfrac{5^\circ F-160}{9}$

If Habib's recipe is to be baked at 356 degree Fahrenheit, then we substitute F=356 to obtain an equivalent value in degree Celsius.

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Habib should bake his scones at $180^\circ C$ .

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2 years ago

Which of the following is the expansion of (3c + d2)6?

vovangra [49]

Answer: Option A) $729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$ is the correct expansion.

Explanation:

on applying binomial theorem, $(a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r$

Here a=3c, $b=d^2$ and n=6,

Thus, $(3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r$

⇒ $(3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6$

⇒ $(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}$

⇒ $(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}$

⇒ $(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$

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