What side is adjacent to angle Y? XZ ZY YX None of the above

I have four questions please answer all! With data/evidence

SOVA2 [1]

To answer all these questions you will use the formula A= bh. In words this means to find the area you multiply the base times the height. For each of these examples, you are given the area and you have to find the missing dimension. Each of these will be set up as four separate equations that need solved. Here they are 0.225=0.6h, 4.86=1.8b, 63=12b, and 2.5=5h. To solve all four of these and show evidence, you were divided the area by the dimension that you know. 0.225÷0.6, 4.86÷1.8, 63÷12, and 2.5÷5. With any multiplication equation if you know the total, you have to divide to find the missing part. The answers for each of these in order that they are presented our 3.75 miles, 2.7 yards, 5.25 m, and 0.5 km. None of the answers are square units because they are only giving a one dimensional measurement.

7 0

4 years ago

Which is the simplified form of (9c^-9)^-3 b 1/729 c^27

rosijanka [135]

ANSWER

$(9 {c}^{ - 9} )^{ - 3}=\frac{{c}^{ 27} }{729}$

EXPLANATION

The given expression is

$(9 {c}^{ - 9} )^{ - 3}$

Recall that:

$({a}^{m} )^{n} = {a}^{mn}$

We use the law of exponents to get

$(9 {c}^{ - 9} )^{ - 3} = 9^{ - 3} {c}^{ - 9 \times - 3}$

Let us multiply in the exponents to get:

$(9 {c}^{ - 9} )^{ - 3} = 9^{ - 3} {c}^{ 27}$

Recall again that:

${a}^{ - m} = \frac{1}{ {a}^{m} }$

This implies that:

$(9 {c}^{ - 9} )^{ - 3} = \frac{1}{ {9}^{ 3} } \times {c}^{ 27}$

We expand the power to get:

$(9 {c}^{ - 9} )^{ - 3} = \frac{1}{9 \times 9 \times 9} \times {c}^{ 27}$

$(9 {c}^{ - 9} )^{ - 3}=\frac{1}{729} \times {c}^{ 27}$

We multiply out to get:

$(9 {c}^{ - 9} )^{ - 3}=\frac{{c}^{ 27} }{729}$

7 0

3 years ago

Rewrite as a simplified fraction 2.67 repeating

Morgarella [4.7K]

265/99, is what I think but I’m not sure

5 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.