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3 years ago

8

I need answers to all of these. Please help!! I suck at math..

1 answer:

RSB [31]3 years ago

6 0

1)  (-4) + (-9) = -13

2) 5 + (-12) = -7

3) -6 + 5 = -1

4) -7 + 7 = 0

<span>5) -7 + 2 ___ 3 + (-9)

- 5 ____ - 6

</span>-7 + 2 >  3 + (-9)

6)  <span>4 + (-8) ___ 6 + (-8) + (-6)

- 4 ____ - 8

</span><span>4 + (-8) >  6 + (-8) + (-6)

7)  </span><span>$66

8)  </span><span>-$12
</span>

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(6,3*10^-2)÷(2,1*10^4

nexus9112 [7]

Answer:

$3*10^{-6}$ or $0.000003$

Given expression:

$(6.3*10^{-2}) : (2.1*10^4)$

Simplify it in steps below:

$(6.3*10^{-2}) : (2.1*10^4) =$
$(6.3 :2.1)*(10^{-2}:10^4)=$
$3*(10^{-2-4})=$
$3*10^{-6}$ or $0.000003$

<u>Used properties:</u>

$\cfrac{ab}{cd}=\cfrac{a}{c}*\cfrac{b}{d}$

$a^b:a^c=a^{b-c}$

$a^{-b}=1/a^b$

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1 year ago

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The length of Suez Canal on a map of drawing scale 1 : 1400000 is 70 Km ,then its

lapo4ka [179]

Answer:

76

Step-by-step explanation:because its the answer

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2 years ago

Can someone please help I’m not so good at fractions

satela [25.4K]

Answer:

A) 25/33

B) 36

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d) 2 11/12

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2 years ago

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Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

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2 years ago

Need help po, needed today. Thanks!

Dimas [21]

Answer:

<h2>question is not up down this is incorrect questions</h2>

8 0

2 years ago