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laila [671]
4 years ago
13

Given that,

Mathematics
1 answer:
Mrac [35]4 years ago
3 0

Answer:

0.6990

Step-by-step explanation:

We have:

$log_{10} (10)=1$

$\therefore \log_{10}(2\times 5)=1$

$\implies \log_{10}(2)+ \log_{10}(5)=1$

$\implies \log_{10}(5)=1-\log_{10}(2)$

$\implies \log_{10}(5)=1-0.3010=\boxed{0.6990}$

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A cube has a volume 7^3 of cubic centimeters. What is its surface area?
kakasveta [241]

Answer:

294 cm^2

Step-by-step explanation:

length of the cube is 7cm each side

S.A = 2(7 * 7) + 2( 7 * 7) + 2( 7 * 7)

= 294 cm^2

4 0
3 years ago
- x2 +10x-16<br> As a graph
Ira Lisetskai [31]
The zero for -2x+10x-16 is positive 2

3 0
3 years ago
For every 50 emails that Nick sent in a month, he received 30 emails. What is the ratio in simplest form of the number of emails
Anna35 [415]
The answer is D. Because, for every 50 emails dent he received 30 back in response. So, let 5 represent the emails he sent and let 3 represent the emails received... so your answer should be D: 5:3.

I hope this helps!!! Have a good day!!! :D
6 0
3 years ago
Read 2 more answers
A delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is 70 p
nikklg [1K]
X = large boxes and y = small boxes

x + y = 70.....x = 70 - y
60x + 65y = 4300

60(70 - y) + 65y = 4300
4200 - 60y + 65y = 4300
5y = 4300 - 4200
5y = 100
y = 100/5
y = 20 <=== the small boxes weigh 20 lbs

x + y = 70
x + 20 = 70
x = 70 - 20
x = 50 <== and the large boxes weigh 50 lbs
5 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
hram777 [196]

Write the system in augmented-matrix form:

\left[\begin{array}{ccc|c}2&2&4&16\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]

Multiply through row 1 by 1/2:

\left[\begin{array}{ccc|c}1&1&2&8\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]

Add -1(row 1) to row 3, and add -5(row 1) to row 2:

\left[\begin{array}{ccc|c}1&1&2&8\\0&-7&-7&-41\\0&1&-5&-17\end{array}\right]

Swap rows 2 and 3:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&-7&-7&-41\end{array}\right]

Add -7(row 2) to row 3:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&-42&-160\end{array}\right]

Multiply through row 3 by -1/42:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&1&\frac{80}{21}\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]

Add -1(row 2) and -2(row3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-\frac53\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]

So the solution to the system is

x=-\dfrac53,y=\dfrac{43}{21},z=\dfrac{80}{21}

7 0
4 years ago
Read 2 more answers
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