Ask question

Ask question

All categories

3 years ago

11

Kelsy climbs up a volcano. When she is a quarter of the way up the volcano she is at a height of 125 m. How high is Kelsy from t

he bottom when she is 75 m below the top of the volcano?

2 answers:

cestrela7 [59]3 years ago

8 0

To find the height, multiply 125 by 4 and subtract 75 to find 75m below the top
125 x 4-75=425m

Leviafan [203]3 years ago

3 0

She is 425 feet above the sea level, or the starting level.

You might be interested in

Ln a+3 + ln a-3 = ln 16

Lilit [14]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)\\\\
and\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
ln(a+3)+ln(a-3)=ln(16)\implies ln[(a+3)(a-3)]=ln(16)
\\\\\\
ln[a^2-3^2]=ln(16)\impliedby \textit{removing ln() from both sides}
\\\\\\
a^2-9=16\implies a^2=16+9\implies a^2=25
\\\\\\
a=\pm\sqrt{25}\implies a=\pm 5$

7 0

3 years ago

A line has a slope of - 4 and includes the points (6,v) and (3,3). What is the value of v?

labwork [276]

Answer: v = - 9

Step-by-step explanation:

Since two points on the line are given, we use formula for slope.

Let (a,b) and (c,d) be two points on the line, the slope of the line is given by

m= $\frac{d-b}{c-a}$ .

Hence, using this we have

-4= $\frac{3-v}{3-6}$

Thus, we have

-4×(-3)=3-v⇒12=3-v⇒12-3=-v⇒v=-9

3 0

4 years ago

A tv programme begins at 7:45 and ends 50 minutes later what time does it finish

faust18 [17]

Answer:

8:35

Step-by-step explanation:

7:45 + 50minutes=

8:35

7 0

3 years ago

Read 2 more answers

If a right triangle's hypotenuse is 17 units long, and one of its legs is 15 units long, how long is the other leg?

Vesna [10]

Answer:

8 units

Step-by-step explanation:

Hello!

So, there's a formula we can apply to right-angled triangles: Pythagorean's theorem. It states that c = $\sqrt{{a}^2 + b^{2} }$ , where <em>c</em> is the hypotenuse and <em>a</em> and <em>b </em>are the legs of the triangle.

So, from the problem, if <em>c </em>= 17 and <em>a </em> = 15, then, we're solving for <em>b</em>. So we'll rewrite the theorem to solve for <em>b</em>.

${c}^2 = {a}^2+{b}^2\\{c}^2-{a}^2={b}^2\\{b} = \sqrt{{c}^2-{a}^2}$

Okay, so now we have isolated the theorem for <em>b. Let's </em>plug in our values for <em>c </em>and <em>a</em>.

$b = \sqrt{{17}^2-{15}^2}\\b = \sqrt{289-225}\\b = \sqrt{64}\\b = 8$

So, using the theorem, we found <em>b</em> = 8. To check our work, let's plug in <em>b</em> and <em>a</em> and solve for <em>c.</em>

<em /> $c = \sqrt{{a}^2+{b}^2}}\\c = \sqrt{{15}^2+{8}^2}\\c = \sqrt{225+64}\\c = \sqrt{289}\\c = 17\\$ <em />

So, we got our hypotenuse to equal 17 units, which is correct! So, our <em>b</em> is correct too. Awesome

7 0

3 years ago

2) Jimmy was going to sell all of his stamp

PSYCHO15rus [73]

48 stamps because 29-5=24 , 24x2=48

8 0

3 years ago