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makvit [3.9K]
3 years ago
7

Write and solve an inequality to find the values of x for which the perimeter of the rectangle is less

Mathematics
1 answer:
avanturin [10]3 years ago
7 0

Answer:

B (if that sign is x + 4)

C (if that sign is x ÷ 4)

Step-by-step explanation:

We know that x is definitely less than the other value, and cannot possibly be equal to the other value, as it's x + 4, so the answer is B.

It's the other way round if it's x ÷ 4 and the answer will be C.

hopefully I interpreted this question the right way!

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I need help please! I've been stuck on this for a week and its really stressful lol
worty [1.4K]

Answer:

y=0.2*4^{x}

Step-by-step explanation:

Notice when x increases 1, y is 4 times the previous one, so

the function is like y=C*4^{x}

To determine the constant C, put any pair of (x, y)

Use x = 0, y = 0.2, so

0.2 = C*4^{0} = C * 1 = C

then y=0.2*4^{x}

5 0
2 years ago
Read 2 more answers
Can someone please help
Rashid [163]

Answer:

It shifted 6 units down

Step-by-step explanation:

Do the graph normally without the k and compare them side by side.

8 0
2 years ago
A cylindrical barrel with a radius of 12 feet rests on its round side against a wall. A ladder leans against both the barrel and
frosja888 [35]
The answer is 24 cool
5 0
3 years ago
Please answer (-4) + (-5)
sertanlavr [38]
-9

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7 0
3 years ago
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
3 years ago
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