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3 years ago

The length is _ the width is and the area of the actual room is _ _______

Mathematics

1 answer:

RSB [31]3 years ago

8 0

Answer:

98 square inches.

Step-by-step explanation:

The length is 14 inches. The width is 7 inches.

7 * 14 = 98 square inches.

The area is 98 square inches.

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Si tenemos un rectángulo en el que su base es igual a x+10 y su altura es igual a x+9, ¿cuál será la expresión algebraica correc

aksik [14]

Responder:

x² + 19x + 90

Explicación paso a paso:

Área del rectángulo = Base * Alto

Dado

Base = x + 10

Altura = x + 9

Área del rectángulo = (x + 10) (x + 9)

Área del rectángulo = x (x) + 9x + 10x + 9 (10)

Área del rectángulo = x² + 9x + 10x + 90

Área del rectángulo = x² + 19x + 90

La expresión que representa el área si la cifra es x² + 19x + 90

8 0

2 years ago

Write two sentences to match this numerical expression:7 times 3

Marina CMI [18]

3 times 7..... . .....

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3 years ago

Read 2 more answers

Susan has been on a bowling team for 14 years. After examining all of her scores over that period of time, she finds that they f

elixir [45]

Answer:

Step-by-step explanation:

Let X be the random variable that represents a score of Susan. We know that X has a normal distribution with average score of 225 and a standard deviation of 13. For n = 16 games, the average score $\bar{X}$ has a normal distribution with mean of 225 and a standard deviation of $13/\sqrt{16}=13/4=3.25$ . Besides, the z-score related to 220 is (220-225)/3.25=-1.5385, and the z-score related to 228 is (228-225)/3.25= 0.9231. Therefore, we are looking the probability $P(220 < \bar{X} < 228) = P(-1.5385 < Z < 0.9231) = P(Z < 0.9231) - P(Z < -1.5385)$ = 0.76

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2 years ago

Can someone help me pls pls help with this question

True [87]

Answer:

Step-by-step explanation:

The mean is 11,000

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2 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

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2 years ago

The length is _____ the width is______ and the area of the actual room is ___ _______

The length is _ the width is and the area of the actual room is _ _______