Solve the equation using square roots. x2 + 10 = 9

A. Plus or minus the square root of 1.

B. no real number solutions

C. 1

D. –1

The measure of angle at the tangent of a circle is a right angle.

Start by calculating the measure of angle ABC using the following angle in a triangle theorem

[tex]\angle ABC + 48 + 47 = 180[/tex]

[tex]\angle ABC + 95 = 180[/tex]

Subtract 95 from both sides of the equation

[tex]\angle ABC = 180 - 95[/tex]

[tex]\angle ABC = 85[/tex]

For line BC to be a tangent, then the following must be true

[tex]\angle ABC = 90[/tex]

Hence, the correct statement about BC is (c)

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Answer:

-1/3

Step-by-step explanation:

We need a c number that:

5/6x + cx = 1/2x

We clear the equation:

cx = 1/2x -5/6x

c = 1/2 -5/6

c= -2/6 = -1/3

Answer:

1.D

2.D

3.B

4.A

5.A

Step-by-step explanation:

1.We are given that two equations

[tex]d+e=15[/tex]

[tex]-d+e=-5[/tex]

Adding two equations then we get

[tex]2 e=10[/tex]

[tex]e=\frac{10}{2}=5[/tex]

Substitute e=5 in equation one then we get

[tex]5+e=15[/tex]

[tex] d=15-5[/tex]

[tex]d=10[/tex]

Hence, the ordered pair as (10, 5).

Therefore,Option D is true.

2.We are given that two equations

Equation S:[tex]y=x+9[/tex]

Equatin T:[tex]y=2x+1[/tex]

Using substitution method

Substitute the value of y from equation one in equation second then we get

[tex]x+9=2x+1[/tex]

Therefore, option D is true.

3.We are given that two equations

Equation C :[tex]y=6x+9[/tex]

Equation D[tex]:y=6x+2[/tex]

The two equations can be written as

[tex]6x-y+9=0[/tex]

[tex]6x-y+2=0[/tex]

[tex]a_1=6,b_1=-1,c_1=9[/tex]

[tex]a_2=6,b_2=-1,c_2=2[/tex]

[tex]\frac{a_1}{a_2}=\frac{6}{6}=1:1[/tex]

[tex]\frac{b_1}{b_2}=\frac{-1}{-1}=1:1[/tex]

[tex]\frac{c_1}{c_2}=\frac{9}{2}[/tex]

[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}[/tex]

Therefore, system of equations have no solution

Hence, option B is true.

4.We are given that two equations

[tex]2x+y=-4[/tex]

[tex]y=3x+2[/tex]

Using substitution method

Substitute value of y from equation second in equation one

Then we get

[tex]2x+3x=2=-4[/tex]

[tex]5 x=-4-2[/tex]

[tex]5 x=-6[/tex]

[tex]x=-\frac{6}{5}[/tex]

Substitute the value of x in equation second then we get

[tex]y=3\times( -\frac{6}{5})+2[/tex]

[tex]y=\frac{-18+10}{5}[/tex]

[tex]y=-\frac{8}{5}[/tex]

Hence, option A is true.

5.We are given that two equations

[tex]y=-2x+1[/tex]

[tex]6x+2y=22[/tex]

Using substitution method

Substitute value of y from equation one in second equation then we get

[tex]6x+2(-2x+1)=22[/tex]

Hence, option A is true.

Answer:

[tex]\frac{5}{2}[/tex]

Step-by-step explanation:

The slope of a line parallel to b has the exact same slope as b. Therefore, by finding the slope of b, you will find the slope of a line parallel to be as well.

The following formula is used to find m, the slope of the line, using the two given points:

[tex]m=\frac{y_{1}-y_{2} }{x_{1}-x_{2} } \\m=\frac{5--5}{3--1} \\m=\frac{5+5}{3+1} \\m=\frac{10}{4} \\m=\frac{5}{2}[/tex]

The required algebraic expression can be written as [tex](y\div4)-3[/tex].

Given: a statement is [tex]3[/tex] less than the quotient of a number [tex]y[/tex] and [tex]4[/tex].

According to question,

An algebraic expression can be written by use of  [tex]3[/tex] less than the quotient of a number [tex]y[/tex] and [tex]4[/tex].

Here, expression is [tex](y\div4)-3=\frac{y}{4}-3[/tex].

Therefore, the algebraic expression for the statement  [tex]3[/tex] less than the quotient of a number [tex]y[/tex] and [tex]4[/tex] is written as [tex]\frac{y}{4}-3[/tex].

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[tex]2m^2-7m-13=-10\ \ \ |+10\\\\2m^2-7m-3=0[/tex]

[tex]a=2;\ b=-7;\ c=-3\\\\b^2-4ac=(-7)^2-4\cdot2\cdot(-3)=49+24=73\\\\m_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ m_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\m_1=\dfrac{-(-7)-\sqrt{73}}{2\cdot2}=\dfrac{7-\sqrt{73}}{4}\\\\m_2=\dfrac{-(-7)+\sqrt{73}}{2\cdot2}=\dfrac{7+\sqrt{73}}{4}[/tex]

His acceleration is, 2 m/s^2!

Hope this helps! :)

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Answer:

Yes because the relative difference between y-values and x-values is the same no matter what which pairs of (x,y) values you use to calculate it

Step-by-step explanation:

Yes, this set of ordered pairs have been generated by a linear function because the relative difference between y-values and x-values is the same no matter which pairs of (x, y) values you use to calculate it.

The correct option is (B)

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

Given are ordered pairs are: (-3,0) (-2,4)(-1,8)(0,12)(1,16)(2,20).

It can be seen that the there only unit increase in x coordinates and units increase in consecutive y- coordinates.

Now, if we set up the equation of line as:

Here, m =4 and b = 12

So the relationship is linear and represented by the equation

y=4x+12

Hence, the relative difference between y-values and x-values is the same no matter which pairs of (x, y) values you use to calculate it.

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Answer:

(2.25π) yd²

Step-by-step explanation:

The area of a circle is given by the formula:

[tex]\boxed{\text{Area of circle}=\pi r^{2} }[/tex] , where r is the radius.

In this question, the radius is 1.5 yd.

Substitute r= 1.5 into the formula:

Area of circle B

[tex]=\pi (1.5)^2[/tex]

= (2.25π) yd²

Additional:

For more questions on area of circle, do check out the following!

In a 30-60-90 triangle, the hypotenuse is twice the length of the shortest leg. Hence, if the shortest leg is 4, then the hypotenuse is 8.

In a 30-60-90 triangle, the ratios of the sides are specific and constant. The shortest side, the one opposite the 30-degree angle, is considered to be x. The longest side (hypotenuse), directly across from the 90-degree angle, is 2x, and the remaining side is x√3. In this case, the shortest side (x) is indicated to be 4.

So, in this triangle, the hypotenuse (2x) would be 2*4 = 8.

The Pythagorean theorem, as mentioned in the reference content, will also hold, but in the case of this special triangle, the ratios of the sides simplify the process of determining side lengths.

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Answer: The correct option is second, i.e., x>0.

Explanation:

As we know that the domain is the set of all possible inputs. If function is defined as, f(x) then all possible value of x for which the function f(x) is defined is called domain.

In a graph the domain is defined on the x axis and the range of the function is defined on y-axis.

In the given graph the function is defined from x=0 to [tex]x=\infty[/tex] because for [tex]x\leq 0[/tex] the graph is not defined. It means for [tex]x\leq 0[/tex] the function is not defined.

Sicen the graph is defined for all positive values of x, therefore the domain of the function is al real number greater than 0. It can be written as x>0 and the second option is correct.