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3 years ago

13

Is (4,4) a solution of the graphed system of inequalities? Choose 1 answer: Yes No

1 answer:

MissTica3 years ago

3 0

No it is not. They are the wrong points

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Please help I need it now please and thank you

Bingel [31]

Answer:

It would be the first one or second one

Step-by-step explanation:

the answer is up ahead

5 0

2 years ago

Read 2 more answers

Consider the image of EFGH for the translation (x, y) → (x – 3, y + 1). What is the ordered pair of F′?

omeli [17]

Answer:

$F'(-1,3)$

Step-by-step explanation:

The rule for the translation is given as

$(x,y)\rightarrow (x-3,y+1)$

The coordinates of F are (2,2).

We substitute these coordinates into the translation rule to find the coordinates of F'.

$F(2,2)\rightarrow F'(2-3,2+1)$

$F(2,2)\rightarrow F'(-1,3)$

3 0

3 years ago

Find the cosine of ∠W.

d1i1m1o1n [39]

Answer:

ok so u =?? 15 and 17= 2 but cant be cs they 2 4 6 8 10 12 14 16 not 15 or 17so 3

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

Why is Point , Line , and Plane considered the 3 building blocks of geometry??? Please answer fast

tangare [24]

These 3 objects are used to make all of the other objects that we will use in geometry

3 0

3 years ago