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Eddi Din [679]
3 years ago
7

Which transformations is NOT an isometry?

Mathematics
1 answer:
BigorU [14]3 years ago
5 0

Answer:

Dilation

Step-by-step explanation: im doing this in k-12 in math now

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Please help me i need it right now please
12345 [234]

Given:

The rate of interest on three accounts are 7%, 8%, 9%.

She has twice as much money invested at 8% as she does in 7%.

She has three times as much at 9% as she has at 7%.

Total interest for the year is $150.

To find:

Amount invested on each rate.

Solution:

Let x be the amount invested at 7%. Then,

The amount invested at 8% = 2x

The amount invested at 9% = 3x

Total interest for the year is $150.

x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150

Multiply both sides by 100.

7x+16x+27x=15000

50x=15000

Divide both sides by 50.

x=\dfrac{15000}{50}

x=300

The amount invested at 7% is 300.

The amount invested at 8% is

2(300)=600

The amount invested at 9% is

3(300)=900

Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.

4 0
3 years ago
List the elements in the set.<br> 1) {x 1 x is an integer between -7 and -3}
omeli [17]

Answer:

सामाजिक अध्ययन र जिवनोपयोगी शिक्षाको महत्व दसांउनुहोस

8 0
1 year ago
Value of x in this? Having trouble with this problem.
Solnce55 [7]
Answer:

I think it’s 105

Step by step explanation:

I hope this helped in some kind of way.
6 0
3 years ago
IM just a little bit slow
viva [34]

Answer:

71°

Step-by-step explanation:

31-3x+19x-5=90

16x=90-26

16x=64

x=4

m<R=19x-5 = 19(4)-5 = 76-5 = 71

5 0
3 years ago
Read 2 more answers
What is the value of log 43? Use the calculator. Round your answer to the nearest tenth.
Bad White [126]

Answer:

The value of log  43is 1.633

Step-by-step explanation:

Explanation:

Suppose you know that:

\begin{array}{l}{\log 2 \approx 0.30103} \\{\log 3 \approx 0.47712}\end{array}

Then note that:

43=\frac{129}{3} \approx \frac{128}{3}=\frac{2^{7}}{3}

So

\log 43 \approx \log \left(\frac{2^{7}}{3}\right)=7 \log 2-\log 3 \approx 7 \cdot 0.30103-0.47712=1.63009

We know that the error is approximately:

\log \left(\frac{129}{128}\right)=\log 1.0078125=\frac{\ln 1.0078125}{\ln 10} \approx 0.00782 .3=0.0034

So we can confidently give the approximation:

\log 43 \approx 1.633

6 0
3 years ago
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