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3 years ago

Find the cosine of ∠J.

Mathematics

1 answer:

Luden [163]3 years ago

7 0

Answer:

$\cos = \frac{base}{hypotenuse} = \\ \cos(j) = \frac{ \sqrt{29} }{ \sqrt{94} } = \sqrt{ \frac{29}{94} } \\ answer = \sqrt{ \frac{29}{94} }$

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Lucas opened a savings account 20 years ago with a deposit of $3,815.24. The account has an interest rate of 3.3% compounded

qwelly [4]

Answer:

Interest earned = $3546.50

Option B is correct option.

Step-by-step explanation:

Time t = 20 years

Principal Amount P =$3815.24

Interest r = 3.3% or 0.033

Compounded quarterly n = 4

We need to find How much interest has Lucas earned?

First we need to find the future value in the account after 20 years

The formula used to find future value A is: $A=P(1+\frac{r}{n})^{nt}$

Putting values of t, r , n and P to find A i.e future value

$A=P(1+\frac{r}{n})^{nt}\\A=3815.24(1+\frac{0.033}{4})^{20*4} \\Simplifying:\\A=3815.24(1+0.00825)^{20*4} \\A=3815.24(1.00825)^{80}\\A=3815.24(1.929)\\A= 7361.74$

Now finding interest amount by subtracting the initial deposit from the future value

Interest earned = Future Value (A)- Initial Deposit (P)

Interest earned = 7,361.74 - 3815.24

Interest earned = 3546.50

So, Lucas have earned interest = $3546.50

Option B is correct option.

3 0

3 years ago

Help me find the value of x

iris [78.8K]

X=22

(X+1)+ (3x+1)= 90

4x+2=90
4x=88
x=22

6 0

3 years ago

I need help with this problem please help.

Nutka1998 [239]

The answer should be SAS, since you know two sides and an angle in the boy’s case, and a side and an angle in the tree’s case.

8 0

2 years ago

What is the radius of a circle with the equation (x − 3)2 + (y − 7)2 = 9? A) 3 B) 4.5 C) 6 D) 9

makvit [3.9K]

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-\stackrel{h}{3})^2+(y-\stackrel{k}{7})^2=9\implies (x-\stackrel{h}{3})^2+(y-\stackrel{k}{7})^2=\stackrel{r}{3^2}~\hfill \begin{cases} center~(3,7)\\ radius=3 \end{cases}$

7 0

3 years ago

Read 2 more answers

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