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Norma-Jean [14]
3 years ago
9

Graph the equations to solve the system y=-2×-7

Mathematics
1 answer:
maxonik [38]3 years ago
3 0
To find this, you insert either [tex]y=0[\tex] or [tex]x=0[\tex], so the line intercepts the y axis on -7, and the x axis on -3.5. We get the following:

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11111nata11111 [884]
The answer is b. The equivalent of -7/8 is 7/-8.
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Show me the greater sign
Amiraneli [1.4K]
The grater sign is >
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I don’t understand how to do it ?
Temka [501]

We know that Sum of Angles in a Triangle is Equal to 180°

Here EBF is a Triangle

⇒ m∠EBF + m∠BEF + m∠EFB = 180°

⇒ 60° + 40° + m∠EFB = 180°

⇒ 100° + m∠EFB = 180°

⇒ m∠EFB = 180° - 100°

⇒ m∠EFB = 80°

As Line m and Line p are Parallel Lines :

Alternate Interior Angles are Equal, here Alternate Interior Angles are m∠BEF and m∠ABE

⇒ m∠BEF = m∠ABE

⇒ m∠ABE = 40°

We know that Vertically Opposite Angles are Equal, Here m∠GFI and m∠EFB are Vertically Opposite Angles.

⇒ m∠GFI = m∠EFB

⇒ m∠GFI = 80°

We can notice that m∠DEB and m∠BEF form a Linear Pair

⇒ m∠DEB + m∠BEF = 180°

⇒ m∠DEB + 40° = 180°

⇒ m∠DEB = 180° - 40°

⇒ m∠DEB = 140°

We can notice that Sum of Angles m∠CBF and m∠EBF and m∠ABE is 180°

⇒ m∠CBF + m∠EBF + m∠ABE = 180°

⇒ m∠CBF + 60° + 40° = 180°

⇒ m∠CBF + 100° = 180°

⇒ m∠CBF = 180° - 100°

⇒ m∠CBF = 80°

We can notice that m∠BFG and m∠EFB form a Linear Pair

⇒ m∠BFG + m∠EFB = 180°

⇒ m∠BFG + 80° = 180°

⇒ m∠BFG = 180° - 80°

⇒ m∠BFG = 100°

We know that Vertically Opposite Angles are Equal, Here m∠BFG and m∠IFE are Vertically Opposite Angles.

⇒ m∠BFG = m∠IFE

⇒ m∠IFE = 100°

3 0
3 years ago
The temperature was 5º. It dropped 3º during the next hour.<br> What was the temperature then?
Nana76 [90]

Answer:

the answer is  2 degrees hope this helps :)

Step-by-step explanation:

6 0
3 years ago
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Solve the following quadratic equation using the quadratic formula. Separate multiple answers with a comma if necessary.
Valentin [98]

Answer:

y^2 -4y +6=0

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

Where a = 1, b= -4 ,c =6

And replacing we got:

y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}

And solving we got:

y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i

Where i =\sqrt{-1}

And the possible solutions are:

y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i

Step-by-step explanation:

For this case we use the equation given by the image and we have:

-y^2 +4y -6=0

We can rewrite the last expression like this if we multiply both sides of the equation by -1.

y^2 -4y +6=0

Now we can use the quadratic formula given by:

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

Where a = 1, b= -4 ,c =6

And replacing we got:

y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}

And solving we got:

y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i

Where i =\sqrt{-1}

And the possible solutions are:

y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i

4 0
3 years ago
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