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3 years ago

Simplify the expression: 15w-6w+14w^2

Mathematics

2 answers:

yan [13]3 years ago

6 0

Add together all like terms (15w,6w)
15w-6w= 9w
So your answer would be
9w+14w²

Darina [25.2K]3 years ago

5 0

=14w2+9w hope this helps

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4) Abe, Barb, Carl, Dee, and Earl all work at Thrifty's. Abe works every day,

Doss [256]

Answer:

Every 60 days.

Step-by-step explanation:

This is very similar to a sequence.

try starting out with:

Abe: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21

Barb: 2, 4, 6, 8, 10

Carl: 3, 6, 9, 12

Dee: 4,8, 12, 16, 20, 24, 28, 32, 36

Earl: 5, 10, ....

so If today they all worked together....

We need a common multiple of 5, 4, 3.... so it is 60

It appears that it will be in 60 days

4 0

3 years ago

Tìm m để phương trình có 1 nghiệm duy nhất x+2y-2z=2 3x+7y-z=5 2x+4y+mz=7

RSB [31]

Rewrite the system of equations in matrix form.

$\begin{bmatrix}1&2&-2\\3&7&-1\\2&4&m\end{bmatrix} \mathbf x = \mathbf b$

This system has a unique solution $\mathbf x = \mathbf A^{-1}\mathbf b$ so long as the inverse of the coefficient matrix $\mathbf A$ exists. This is the case if the determinant is not zero.

We have

$\det(\mathbf A) = m+4$

so the inverse, and hence a unique solution to the system of equations, exists as long as m ≠ -4.

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2 years ago

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=10

Tju [1.3M]

Answer: 0.8238

Step-by-step explanation:

Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with $\mu=106$ and $\sigma=15$ .

Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.

Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-

$P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]$

Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238

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Answer:

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