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2 years ago

Francine recorded today's high

Mathematics

1 answer:

fenix001 [56]2 years ago

7 0

Answer:

3.5

Step-by-step explanation:

the problem says 4.7 less than which refers to subtraction so we can plug our numbers into a subtraction problem like so

78.6-4.7 which equals 73.9

the second part of the problem asks the difference between yesterday's low and today's.

we can take the two lows and subtraction them to find this answer.

77.4-73.9 which equals 3.5

so today's low was 3.5 degrees cooler than yesturdays

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Sarah is going to invest $280 and leave it in an account for 9 years. Assuming the

Alisiya [41]

Answer:

4.2%

Step-by-step explanation:

5 0

2 years ago

What is m<KNL? <br>Enter your answer in the box<br>

steposvetlana [31]

The measurement of ∠KNL is 102°

First, you must set up an equation to find x.

$5x+2=[3(x+14)$

Then, you must simplify both sides of the equation.

$5x+2=3x+42$

Next, subtract 3x on both sides.

$2x+2=42$

Then, subtract 2 on each side.

$2x=40$

Next, divide 2 on each side.

$x=20$

Now that we have the value of x, we can substitute it in for angle KNL.

$[3(20+14)]$

$[3(34)]$

$102$

$102$ °

3 0

2 years ago

What is the following product?

KiRa [710]

ANSWER

$2 \sqrt{42} +7 \sqrt{2}- 6- \sqrt{21}$

EXPLANATION

The given product is:

$( \sqrt{14} - \sqrt{3} )( \sqrt{12} + \sqrt{7} )$

We expand using the distributive property to obtain:

$\sqrt{14}( \sqrt{12} + \sqrt{7}) -\sqrt{3}( \sqrt{12} + \sqrt{7} )$

Extract the perfect squares to get:

$\sqrt{14}(2 \sqrt{3} + \sqrt{7}) -\sqrt{3}( 2\sqrt{3} + \sqrt{7} )$

Expand further to get;

$2 \sqrt{42} +7 \sqrt{2}- 2(3) - \sqrt{21}$

This simplifies to,

$2 \sqrt{42} +7 \sqrt{2}- 6- \sqrt{21}$

5 0

2 years ago

PLEASE HELP ME!!!!!!! Find the volume of the shape below. 4m 10m 4m 3m

Alex787 [66]

Answer:

the answer is 480 all u have to do is multiplie all the nombers together

5 0

3 years ago

Please help!!!<br> Which expression is equivalent to (3m^-1n^2)^4/(2m^-2n)^3

Elza [17]

$\text{Use:}\\\\(a\cdot b)^n=a^nb^n\\\\(a^n)^m=a^{n\cdot m}\\\\\dfrac{a^n}{a^m}=a^{n-m}$

$\dfrac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}=\dfrac{3^4(m^{-1})^4(n^2)^4}{2^3(m^{-2})^3n^3}=\dfrac{81m^{-4}n^8}{8m^{-6}n^3}\\\\=10.125m^{-4-(-6)}n^{8-3}=10.125m^2n^5$

7 0

2 years ago

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