At a cafe, for every 5 sodas sold there are 9 teas sold. whay is the ratio of teas sold to sodas sold?

A submarine lies 3 kilometers from the path of a target ship. The sub’s torpedoes have a range of 5 kilometers. For what distanc

aev [14]

The distance along the path of the ship be within the range of the torpedoes is 4 kilometers.

Explanation:

The submarine lies 3 km from the target ship.

Let B denotes the target ship and C denotes the submarine.

Let AC denotes the the target ship would be within the range of the torpedoes.

The image attached below illustrates that the distance between submarine and the target ship and the range of the torpedoes.

Let BC denotes the distance between the target ship and the submarine. Let AC denotes the range of the torpedoes.

To determine the length of AB, let us use Pythagorean theorem,

$AC^{2} =AB^{2} +BC^{2}$

Substituting the values of BC = 3 and AC = 5, we have,

$5^{2} =AB^{2} +3^{2}$

Squaring the terms, we get,

$25 =AB^{2} +9$

Subtracting both sides by 9, we have,

$16 =AB^{2}$

Taking square root on both sides,

$AB = 4$

Thus, the distance along the path of the ship be within the range of the torpedoes is 4 kilometers.

5 0

4 years ago

I really need the help please and thank you

Umnica [9.8K]

Answer:

B. $f^{-1}(x)=\frac{x-4}{5}$

Step-by-step explanation:

To find the inverse of a function/equation:

Replace f(x) with y: y=5x+4
Switch x and y: x=5y+4
Solve for y: x-4=5y -> $y=\frac{x-4}{5}$
Write in standard inverse notation: $f^{-1}(x)=\frac{x-4}{5}$

I hope this helps!!

gl!

3 0

3 years ago

Suppose the time that it takes a certain large bank to approve a home loan is Normally distributed with mean (in days) μ and sta

kolezko [41]

Answer:

(c) H0 should be rejected

Step-by-step explanation:

Null hypothesis (H0): population mean is equal to 5

Alternate hypothesis (Ha): population mean is greater than 5

Z = (sample mean - population mean) ÷ (sd/√n)

sample mean = 5.3, population mean = 5, sd = 1, n = 500

Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67

Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326

Conclusion:

Since 6.67 is greater than 2.326, reject the null hypothesis (H0)

4 0

3 years ago

What is the coefficient of abc when the product (a + 2b)(b + 2c)(c + 2a) is expanded and

oksian1 [2.3K]

Answer:

The coefficient is 9

Step-by-step explanation:

(a)(b)(c)+(a)(2c)(2a)+(2b)(b)(c)+(2b)(2c)(2a)

abc+4ca^2+2cb^2+8abc

9abc+4ca^2+2cb^2

8 0

3 years ago

If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago