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pickupchik [31]
3 years ago
11

-45+(-30)-90 the answer of that mathematics sum

Mathematics
2 answers:
Klio2033 [76]3 years ago
6 0
The answer of
<span>-45+(-30)-90= -165

Hope this helps <3
</span> 
kolbaska11 [484]3 years ago
6 0
The answer to your question is 1260.
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What kind of property is this problem 3x(k+4)=3xk+12
ser-zykov [4K]
If i'm not wrong its distributing property of addition. <span />
4 0
3 years ago
Solve 3x + 2 = 15 for x using the change of base formula log base b of y equals log y over log b. −1.594 0.465 2.406 4.465
disa [49]

<u>Answer:</u>

The value in 3x + 2 = 15 for x using the change of base formula is 0.465 approximately and second option is correct one.

<u>Solution:</u>

Given, expression is 3^{(x+2)}=15

We have to solve the above expression using change of base formula which is given as

\log _{b} a=\frac{\log a}{\log b}

Now, let us first apply logarithm for the given expression.

Then given expression turns into as, x+2=\log _{3} 15

By using change of base formula,

x+2=\frac{\log _{10} 15}{\log _{10} 3}

x + 2 = 2.4649

x = 2.4649 – 2  = 0.4649

Hence, the value of x is 0.465 approximately and second option is correct one.

3 0
3 years ago
Read 2 more answers
Jeremy went school supply shopping and bought a calculator, pencils, markers, and notebooks for $17.50.
faltersainse [42]

Answer:

$14

Step-by-step explanation:

x/17.5=20/100, do the butterfly method

17.5x20=350

350/100= 3.5

17.5-3.5=14

8 0
3 years ago
Read 2 more answers
Jenna earned an 80% on her Benchmark Test. If she answered 36 questions correctly, how many questions were on the test?​
spin [16.1K]
There were 45 questions on the test. Hope this helps! <3
3 0
3 years ago
how many gallons of a 50 antifreeze solution must be mixed with 90 gallons of 20% antifreeze to get a mixture that is 40% antifr
Tamiku [17]
Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.
3 0
3 years ago
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