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pochemuha
3 years ago
10

X+1 ----- ≥ 2 3 Solve for what x represents

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
5 0
\frac{x+1}{3} \geq 2

times 3 both sides
x+1≥6
minus 1
x≥5

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PLEASE HELPPPPPPPPPP!!!! AI HAVE 10 MINUTES LEFTTTTT
Lemur [1.5K]

The formula is in a quadratic formula shape: y = ax^2+bx+c

The function is negative since the value 'a' the coefficient of x^2 is negative meaning that the quadratic function is facing downards. If you don't get this explanation, take a look at the diagram I uploaded.

That means the highest point of the graph is its vertex.

But we must find where the vertex is at or the 't' value.

   't' value at which vertex is = -\frac{b}{2a} =-\frac{16}{2*(-16)} =\frac{1}{2}

Now plug this 't' value into h(t) to find the maximum height, and we will get 36 feet.

Hope that helps!

4 0
3 years ago
Which expression is a perfect cube?
ycow [4]

Answer:

y^24

Step-by-step explanation:

a variable with an exponent has an exponent which is divisible by 3 then it is a perfect cube.

3 0
3 years ago
Read 2 more answers
Please help me 16% of what is 41?
saul85 [17]

The answer is 256.25


EXPLANATION


Let x represent the number.



Then, \frac{16}{100}x=41



We cross multiply yo obtain,



16x=41\times 100



We now divide both sides by 16.



x=\frac{4100}{16}



This will give us,


x=256.25




Therefore the number is 256.26

8 0
3 years ago
Let f(x)=x^2+8x. Evaluate f(−4)=
nevsk [136]

Answer:

f(-4) = - 16

Step-by-step explanation:

f(-4) = (-4)2 + 8(-4)

f(-4) = 16 + -32

f(-4) = - 16

3 0
3 years ago
Read 2 more answers
Find the perimeter of the triangle whose vertices are the following, specified points in the plane.
hichkok12 [17]

Given:

The vertices of the triangle are (-10,-3), (1,4) and (-1,7).

To find:

The perimeter of the triangle.

Solution:

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the vertices of the triangle are A(-10,-3), B(1,4) and C(-1,7).

Using distance formula, we get

AB=\sqrt{(1-(-10))^2+(4-(-3))^2}

AB=\sqrt{(1+10)^2+(4+3)^2}

AB=\sqrt{(11)^2+(7)^2}

AB=\sqrt{121+49}

AB=\sqrt{170}

Similarly,

BC=\sqrt{\left(-1-1\right)^2+\left(7-4\right)^2}=\sqrt{13}

AC=\sqrt{\left(-1-\left(-10\right)\right)^2+\left(7-\left(-3\right)\right)^2} =\sqrt{181}

Now, the perimeter of the triangle is

Perimeter=AB+BC+AC

Perimeter=\sqrt{170}+\sqrt{13}+\sqrt{181}

Perimeter\approx 13.038+3.606+13.454

Perimeter=30.098

Therefore, the perimeter of the triangle is 30.098 units.

8 0
3 years ago
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