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3 years ago

How many numbers are 10 units from zero on number line

Mathematics

2 answers:

Neko [114]3 years ago

6 0

Answer:

Step-by-step explanation:

The reason I said 2 is becasue you could go left and right. To the left of 0 is negitive numbers and to the right of 0 id positive numbers. So 10 units to the right of 0 is 10. Ten units to the left of zero is -10.

Hopefuly this helped you! Please tell me if I am Incorrect. That would be a great help.

Lelu [443]3 years ago

4 0

Answer:

I think 2.

Step-by-step explanation:

10, and negative ten

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Solve the equation with quadratic formula<br> y^2+3y-10=0

neonofarm [45]

Answer:

y = 2 and y = -5

Step-by-step explanation:

Because the coefficient is not greater than 1, you can simply solve the problem without the quadratic formula.

By using the "x" method, you must figure out what multiplies to give you -10, and adds to give you 3

So, you find your values as 5 and -2. You then set these values equal to zero in terms of Y. As so---- (y+5)=0 and (y-2)=0

Then, you solve out to get your answer! Y is equal to -5 and 2

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3 years ago

m∠3 is (3x + 4)° and m∠5 is (2x + 11)°. Angles 3 and 5 are . The equation can be used to solve for x. m∠5 = °

ratelena [41]

I just did this question!

Answers:

1) <em>same side interior angles</em>

2) (3x+4)+(2x+11) = 180

3) 77

7 0

3 years ago

Read 2 more answers

Triangle PQR has vertices P(–2, 6), Q(–8, 4), and R(1, –2). It is translated according to the rule (x, y) → (x – 2, y – 16). Wha

Rudik [331]

The point P has coordinates (x,y) = (-2,6) so x = -2 and y = 6

Replace x and y with those values into the rule given
So,
(x,y) ---> (x-2, y-16)
turns into
(-2,6) ---> (-2-2, 6-16) = (-4,-10)

P = (-2,6)
P ' = (-4,-10)

The answer is -10 because your teacher just wants the y coordinate of point P'

4 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Help me with this problem, i have to simply this ratio

Elena-2011 [213]

Keep on dividing them both by 2

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3 years ago