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SVETLANKA909090 [29]

3 years ago

10

If x+5=11what is the value of x

1 answer:

STALIN [3.7K]3 years ago

8 0

The value of x is 6

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A coordinate grid is shown below: A coordinate grid from negative 2 to 0 to positive 2 is drawn. There are three grid lines betw

Anastasy [175]

Answer:

Part A: 0

Part B: (-1,-1) since point (-1,1) is located in quadrent four when it is reflected over the y axis the second point would be located in quadrent three and the x and y would both be negative.

Step-by-step explanation:

3 0

3 years ago

janice wants to jog 3.25 miles on the treadmill she has jogged 1.63 miles how much farther does she have to jog to meet her goal

Mumz [18]

$3.25 - 1.63 = 1.62$
1.62 miles to go to reach her goal.

4 0

3 years ago

K^2+5k-6=0 quadratic formula

s2008m [1.1K]

I hope this helps you

1.k^2+5k-6=0

a=1

b=5

c= -6

disctirminant =b^2-4ac

disctirminant =5^2-4.1. (-6)

disctirminant =49

x1= -(5)+ square root of 49/2.1

x1=-5+7/2

x1=1

x2= -5- square root of 49/2.1

x2= -5-7/2

x2= -6

7 0

3 years ago

If 3/4 is the dividend and -2/5 is the divisor then what is the quotient ?

Alexxx [7]

Answer:

$\dfrac{-15}{8}$

Step-by-step explanation:

<h3>Fraction division:</h3>

$\sf \dfrac{3}{4} \ \div \dfrac{-2}{5}$

Use KCF method.

Keep the first fraction.
Change division to multiplication
Flip the second fraction.

$\sf \dfrac{3}{4} \ \div \dfrac{-2}{5}=\dfrac{3}{4}*\dfrac{-5}{2}$

$\sf = \dfrac{3*(-5)}{4*2}\\\\ =\dfrac{-15}{8}$

4 0

1 year ago

Prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x^2+y^2=5

zmey [24]

Differentiate both sides of the equation of the circle with respect to $x$ , treating $y=y(x)$ as a function of $x$ :

$x^2+y^2=5\implies2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy$

This gives the slope of any line tangent to the circle at the point $(x,y)$ .

Rewriting the given line in slope-intercept form tells us its slope is

$x=2y+5\implies y=\dfrac12x-\dfrac52\implies\mathrm{slope}=\dfrac12$

In order for this line to be tangent to the circle, it must intersect the circle at the point $(x,y)$ such that

$-\dfrac xy=\dfrac12\implies y=-2x$

In the equation of the circle, we have

$x^2+(-2x)^2=5x^2=5\implies x=\pm1\implies y=\mp2$

If $x=-1$ , then $-1=2y+5\implies y=-3\neq2$ , so we omit this case.

If $x=1$ , then $1=2y+5\implies y=-2$ , as expected. Therefore $x=2y+5$ is a tangent line to the circle $x^2+y^2=5$ at the point (1, -2).

7 0

4 years ago

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