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3 years ago

What is the difference? 2x7 - 8x7

Mathematics

2 answers:

kakasveta [241]3 years ago

7 0

The answer is -42.

Hope it helps.

Tpy6a [65]3 years ago

7 0

= 2x7 - 8x7
multiply first before subtracting
= (2x7) - (8x7)
multiply each set of parentheses
= 14 - 56
the answer will be a negative
= -42

ANSWER: -42

Hope this helps! :)

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

Carlo was baking cupcakes. The recipe called for 1.5 cups of sugar to make 24 cupcakes. How many cups of sugar will he need to m

IceJOKER [234]

Answer:0.75

Step-by-step explanation:12 is half of 24

Answer should be be half of 1.5

3 0

2 years ago

Y = 2x<br> 3x-y=15<br><br> Solve By Substitution

Minchanka [31]

The answer is X=15...

6 0

3 years ago

Read 2 more answers

How do you decide which variable to isolate in order to use the substitution method to solve a system of linear equation?

Kazeer [188]

It doesn’t really matter which variable you isolate first but usually you would use the one that’s by itself already. like for example one of the equations was y = 8. you would already have your y solve for so you would just have to plug that in for y in the other equation. personally, i usually do x first unless one of the equations has either x or y by itself already. i think its easier to just do x first and then solve for y after that, but it just depends on what the equations are; sometimes it might be easier to just do y first. hope this helps!

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2 years ago

Please help as soon as possible

ch4aika [34]

To solve this problem you must apply the proccedure shown below:

You must switch the variables, and then, you must solve for $y$ , as following:

$x=-\frac{1}{2} \sqrt{y} +3\\ (x-3)^{2} = (\frac{1}{2}\sqrt{y})^{2} \\ y=4x^{2} -24x+36\\ f^{-1} (x)=4x^{2} -24x+36$

Therefore, as you can see, the answer is: $f^{-1} (x)= 4x^{2} -24x+36
$

8 0

3 years ago