All categories

3 years ago

Can someone verify her answer please

Mathematics

1 answer:

Kamila [148]3 years ago

7 0

Answer:

Yep, she got it correct.

Step-by-step explanation:

Hope this helped.

You might be interested in

Which of the following mixed numbers is equivalent to 12.1 6 ?

Sliva [168]

Well, you don't have any options.
But a mixed number for 12.16 would be 12 4/25. Hope it helps! :)

7 0

3 years ago

Read 2 more answers

The coordinates below lie on the line with the given slope. find the missing coordinate value. (2,2) (5,y) slope = 2

vodka [1.7K]

Step-by-step explanation:

Hey there!

The given points are; (2,2) and (5,y).

Slope (m) =2

Using formula for slope, we get;

$m = \frac{y2 - y1}{x2 - x1}$

Put all values.

$2 = \frac{y - 2}{5 - 2}$

Simplify it to get value of"y".

$2 = \frac{y - 2}{3}$

$6 = y - 2$

$y = 6 + 2$

Therefore, y= 8.

Hope it helps...

5 0

4 years ago

Forty people were surveyed about their favorite flavor of ice cream. How many like just chocolate? If fifteen like vanilla and n

amm1812

B 16
15+9=24-40= b. 16
hope it helped

3 0

3 years ago

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0

3 years ago

Please I need help !!!

makkiz [27]

Answer:

Step-by-step explanation:

4 and 6

5 0

3 years ago