Is (-2,-6) a solution of y < 7+ + 8?

Slope is the change in Y over the change in x.

Use 2 points on the graph(x,y) to use to solev6.

(2,3) and (0,-7)

Slope = (-7 -3) / (0-2)

Slope = -10/-2

Slope = 5

Answer: A the slope is 5

Check the picture below.

The value of f(x) at x=4 is 2.

A function is an expression, rule, or law in mathematics that describes a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are common in mathematics and are required for the formulation of physical relationships in the sciences.

We have to find the value of function for x=4.

This means that from the figure we have to find the value of y when x= 4.

Now, looking into the figure we have

f(4)= 2

f(0)= -2

f(-1)= -1

f(3)= 4

So, the required value is 2.

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Answer: 8.5

Step-by-step explanation: (42.50 - 17) ÷ (5 - 2)

= 25.50 ÷ 3

= 8.5

Thank me later

Answer:

2

Ok lets start off with your 3 numbers

First do 10 / 5 = 2

Then do 12 / 6 = 2

After that do 20 / 10 = 2

now you see you that its all equivalent to 2 that's your answer!

[tex]\dfrac 25 (d-10) - \dfrac 23(d+6)\\\\\\=\dfrac{6}{15} (d-10) - \dfrac{10}{15} (d+6)\\\\\\=\dfrac{6(d-10) -10(d+6)}{15}\\\\\\=\dfrac{6d -60-10d-60}{15}\\\\\\=\dfrac{-4d-120}{15}\\\\\\=-\dfrac{4d}{15} - 8[/tex]

Step-by-step explanation:

although there is no graph visible here, I suspect this is the same question as I just answered twice already.

the slope is -6.

D is the right answer.

The strategy that won't help her to calculate 9 - 7 is (a) Make 10

The given parameters are:

[tex]Yellow = 7[/tex]

[tex]Red = 9[/tex]

To determine 9 - 7,

However, she cannot make 10.

This is so, because 9 and 7 are not up to 10

Hence, the strategy that won't help her to calculate 9 - 7 is (a)

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Answer:

applied to the motor it reaches a steady-state speed of rotation of 20 rev/s and draws 40 A. What will be (a) the power input to the motor, (b) the power loss in the armature, (c) the torque generated at that speed

voltage of 120 V is applied to the motor it reaches a steady-state speed of rotation of 20 rev/s and draws 40 A. What will be (a) the power input to the motor, (b) the power

Answer:

[tex]{ \rm{V = \frac{lwh}{3} }} \\ \\ { \rm{3V = lwh}} \\ \\ { \boxed{ \rm{ \: w = \frac{3V}{lh } \: }}}[/tex]

4 3/4 = 19/4

133/(19/4) = 133(4/19) = 1197/19 = 63 mph

Answer:

The answer is 4

Step-by-step explanation:

(-2.4)(10) / -3(2)

-24/-6

=4

Answer:

The answer will be 4! Hope this helps

Answer:

16x7 = 112 (16 ounces, 7 hours) 112 ounces to pints = 7 pints of water

Step-by-step explanation:

Answer:

$1650 is the answer hope it helps

[tex]\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}[/tex]

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0  y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

[tex](x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill[/tex]

Answer:

Step-by-step explanation:

-3x + 6y = 11

Put into slope-intercept form to find the slope

6y = 3x + 11

 y = 3x/6 + 11/6

 y = ½x + 11/6

so slope is ½

perpendicular lines have negative reciprocal slopes, so our perpendicular line has slope -2

the slope point form of the perpendicular line will be

(y - 3) = -2(x -(-2))

(y - 3) = -2(x +2)

the slope intercept form will be

y = -2x - 4 + 3

y = -2x - 1

The set of ordered pairs is {(2, 1), (2, 3), (3, 4)} which represents ƒ -1, which is the correct answer that would be an option (C).

An ordered pair is made up of two components isolated by a comma and put inside parentheses. (x, y) indicates an ordered pair, where 'x' is the first component and 'y' is the second component in the ordered pair.

Given the function ƒ = {(1 , 2), (2 , 3), (3 , 4)}

According to option (A),

The set of ordered pairs is {(1, 2), (3, 2), (4, 3)}

This represents ƒ -3,

According to option (B),

The set of ordered pairs is {(2,1), (3, 2), (4, 3)}

This represents ƒ -2,

According to option (C),

The set of ordered pairs is {(2, 1), (2, 3), (3, 4)}

This represents ƒ -1,

Thus, the set of ordered pairs is {(2, 1), (2, 3), (3, 4)} which represents ƒ -1.

Hence, the correct answer would be option (C).

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Answer:

What's the question lol

Answer:

d.

Step-by-step explanation:

[tex]y = x^{\tfrac 1{\ln x}}\\\\\implies \ln y = \ln \left( x^{\tfrac 1{\ln x}} \right)\\\\\implies \ln y = \dfrac{1}{\ln x} \cdot \ln x\\\\\implies \ln y = 1\\\\\implies y =e\\\\\\\text{Now,}\\\\\ \lim \limits_{x \to \infty} x^{\tfrac 1{\ln x}}=\lim \limits_{x \to \infty} y = \lim \limits_{x \to \infty} e = e[/tex]

Answer:

-1.5

Step-by-step explanation:

Answer:

w=4

Step-by-step explanation:

72/48=6/w

3/2=6/w

12=3w

w=4

Answer:

I think this is what is wanted

Step-by-step explanation:

Answer:

<

Step-by-step explanation:

Good Question!

8/20=4/10=2/5=0.4

7/13= lcm of 5 and 13 is 75.

26/75 [<] 35/75

:)

Answer: < less than

Step-by-step explanation:

1. Find LCM of 20 and 13.

20: 20,40,60,80,100,120,140,160,180,200,220,240,260,280

13: 13, 26, 39, 52, 65, 78, 91, 104, 107, 110 ...... 234, 247, 260

The LCM is 260.

2. 20*13 = 260 13*20=260

therefore we need to multiply the numerator by those 13 and 20.

8*13 = 104 7*20 = 140

3. compare Fractions:

(104/260) < (140/260)

[tex]x = 23[/tex]

Step-by-step explanation:

First, we need to write the equations in their standard form, i.e., all the variables and their coefficients are to be placed on the left hand side and the plain numbers on the right hand side. The first equation is already in its standard form:

[tex]2x + 5y = 35\;\;\;\;\;(1)[/tex]

but we need to rewrite the 2nd one. So the standard form of the 2nd equation can be written as

[tex]x + 15y = -10\;\;\;\;\;(2)[/tex]

Now that the equations are in their standard forms, the next step that we need to do is to eliminate one of the variables, namely the y. To do that, we need to multiply either one of the equations so that when we add them together, the y-variable gets eliminated. Note if we multiply Eqn(1) by -3, we'll get -15y on one equation and a +15y on the other equation so that when they are added together, they cancel out. So if we do that, we get

[tex]-6x - 15y = -105\;\;\;\;\;(3)[/tex]

Adding Eqn(3) to Eqn(2), we get

[tex]-5x = -115[/tex]

or

[tex]x = \dfrac{-115}{-5} = 23[/tex]

Formula:

We know that

nth term of the AP is Tn = a+(n-1)d

Where, a = First term

d = Common difference

n = number of terms

Now,

Given that

pth term of the A.P. = Tp = q

⇛a + (p-1)d = q

⇛(p-1)d = q - a

⇛d = (q-a)/(p-1) →→→→(i)

and

qth term of the A.P. = Tq = p

⇛a + (q-1) d = p

⇛ (q-1)d = p-a

⇛ d = (p-a)/(q-1) →→→→(ii)

From Eqn(i) and Eqn(ii)

⇛(q-a)/(p-1) = (p-a)/(q-1)

On applying cross multiplication then

⇛(q-a)×(q-1) = (p-a)×(p-1)

⇛q²-q-aq+a = p²-p-ap+a

⇛q²-q-aq = p²-p-ap

⇛ap-aq = p²-p+q-q²

⇛a(p-q) = (p²-q²)-(p-q)

⇛a(p-q) = (p+q)(p-q)-(p-q)

⇛ a(p-q) = (p-q){(p+q)-1}

On cancelling (p-q) both sides then

⇛a = (p+q-1) →→→→Eqn(iii)

On Substituting the value of a in Eqn(ii) then

⇛ d = [p-(p+q-1)]/(q-1)

⇛d = (p-p-q+1)/(q-1)

⇛d = (-q+1)/(q-1)

⇛d = -(q-1)/(q-1)

⇛d = -1

We have,

a = p+q-1 and d = -1

Now, rth term of the AP

⇛ Tr = a + (r-1)d

⇛Tr = (p+q-1)+(r-1)(-1)

⇛ Tr = p+q-1+(-r+1)

⇛ Tr = p+q-1-r+1

⇛ Tr = p+q-r

Therefore, Tr = p+q-r

Answer: rth term of the given A.P. is p+q-r

Answer:

rth=q+p+qxp

Step-by-step explanation:

rth= 3 of the T and Tq added together and joined.

Answer:

  quotient term: 6

  dividend term: 102

Step-by-step explanation:

[tex]\dfrac{782}{17}=\dfrac{680+102}{17}=\dfrac{680}{17}+\dfrac{\boxed{102}}{17}=40 +\boxed{6} = 46[/tex]