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Elan Coil [88]
2 years ago
13

Rebecca’s bank statement shows a change of -1.50 $ in her account each week how many weeks will it take before the total change

is -$12 ?
Mathematics
1 answer:
White raven [17]2 years ago
7 0
It will take 8 weeks for her bank account to be at -$12
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3 years ago
The cost of one pound of bananas is greater than $0.41 and less than $0.50. Sarah pays $3.40 for x pounds of bananas. Which ineq
vredina [299]

Answer:

0.41<3.40/x<0.50

Step-by-step explanation:

Given that the cost of one pound of bananas is greater than $0.41 and less than $0.50. That is,

If the cost of one banana is P, then, the inequality will be

0.41 < P < 0.50

Sarah pays $3.40 for x pounds of bananas. The inequality that represents the range of possible pounds purchased will be achieved by below

3.40/0.41 = 8.29

3.40/0.50 = 6.8

Therefore, the inequality that represents the range of possible pounds purchased is

6 < x < 9 this is the same as 0.41<3.40/x<0.50

6 0
3 years ago
Plz I need help help me plzzzz
Annette [7]

Answer:

4400

Step-by-step explanation:

15 percent is 660

this means 1 percent is 44

this means 100 percent is 4400

please give me brainliest!<3

5 0
3 years ago
Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.
vitfil [10]

Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152

To find the area of the circumscribed circle:

r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

AC=\sqrt{10.152^2-8^2}\approx 6.25

The area of the triangle is:

A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25

The semiperimeter of the triangle is:

\dfrac{10.152+6.25+8}{2}\approx 24.4

The radius of the circle is therefore \dfrac{25}{24.4}\approx 1.025

The area of the inscribed circle then is \pi\cdot (1.025)^2\approx 3.298.

Hope this helps!

6 0
3 years ago
2log(base 5)^(x-3)-log(base 5)^8= log(base 5)^2
Alja [10]

Answer:

x = 7

Step-by-step explanation:

2log_5(x-3)-log_58= log_52\\\\2log_5(x-3) =log_52+log_58\\\\2log_5(x-3) =log_5(2\times 8) \\\\2log_5(x-3) =log_516\\\\2log_5(x-3) =log_54^2 \\\\2log_5(x-3) =2 log_54 \\\\log_5(x-3) =log_54 \\\\x - 3 = 4\\\\x = 4+3\\\\\huge \orange { \boxed{x = 7}}

7 0
2 years ago
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