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3 years ago

Tell whether x= 5 is a solution of 2x > 76.

Mathematics

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

Step-by-step explanation:

If x = 5, then one side of the equation would be 10, that is not greater than 76

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A recycling center pays 2 cents per aluminum can and 5 cents per glass bottle. The equation 2x+5y=100 describes the number of ca

Liono4ka [1.6K]

Answer:

The x-intercept is 50 and it means to earn $1 using ONLY aluminum cans, you would have to collect 50 of them.

The y-intercept is 20 and it means to earn $1 using ONLY glass bottles, you would have to collect 20 of them.

Step-by-step explanation:

To find x intercept, we set y = 0.

To find y intercept, we set x = 0.

x-intercept:

$2x+5y=100\\2x+5(0)=100\\2x=100\\x=50$

The x-intercept is 50 and it means to earn $1 using ONLY aluminum cans, you would have to collect 50 of them.

y-intercept:

$2x+5y=100\\2(0)+5y=100\\5y=100\\y=20$

The y-intercept is 20 and it means to earn $1 using ONLY glass bottles, you would have to collect 20 of them.

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3 years ago

What is 1/3 as a percentage

almond37 [142]

It would equal about 3.333 % repeating.

7 0

3 years ago

Read 2 more answers

Identify the initial amount and the growth rate of the following: Y = 250 (1+0.2)

docker41 [41]

Answer:

y = 300

Step-by-step explanation:

y = 250 (1+0.2)

First add the 1 and 0.2

y=250(1.2)

multiply 250 with 1.2

y = 300

3 0

3 years ago

Which statement is true about the number of faces on any prism?

antiseptic1488 [7]

B. It is the number of edges on one base plus 2.

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3 years ago

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.

4 0

3 years ago