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dezoksy [38]
3 years ago
13

Please help, i cannot solve.

Mathematics
1 answer:
Ludmilka [50]3 years ago
3 0
Im sorry im in sixgh grade i was so close to solving it but that didnt work, i tried
You might be interested in
Evaluate the expression for a = 5 and b = 12.
Evgesh-ka [11]
The answer is c
(12^2)- (4x5)
=124
5 0
3 years ago
Read 2 more answers
What are the domain and range of f(x)=logx-5?
SCORPION-xisa [38]

Domain ( all possible values of x)  is x > 5    or in interval notation it is  (5, ∞)

The range( all possible values of log x) is  (-∞, ∞)

8 0
3 years ago
Easiest way to calculate 22% mark up from a cost of say 39.99
satela [25.4K]

Answer:

The markup amount on the cost 39.99 $ will be: 8.78 $

Step-by-step explanation:

Given

  • The cost = 39.99 (Let say in $)
  • Mark up = 22%

As

22% = 0.22

The markup of 22% on the cost can be calculated by multiplying the cost by 22% or 0.22

so

Markup amount = 0.22 × 39.99

                           = 8.78 $

Therefore, the markup amount on the cost 39.99 $ will be: 8.78 $

6 0
2 years ago
Mount Kosciuszko is the highest mountain in Australia, with a peak of about 2,228 meters above sea level. Lake Erye has the lowe
marta [7]

Answer:

Option (A)

Step-by-step explanation:

Heights above the sea level are denoted with positive notations and depths below the sea level with negative.

Height of the highest peak of the mountain = +2228 meters

Lowest point of the lake = -15 meters

Difference between these elevations = 2228 - (-15)

                                                              = 2243 meters

Therefore, Option (A) will be the correct option.

6 0
3 years ago
Can someone solve this with steps cause idk how to solve the radicals
IceJOKER [234]
\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9


to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
8 0
3 years ago
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