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3 years ago

Which ordered pair is a solution if the equation? 2x + 3y = 10

Mathematics

1 answer:

adoni [48]3 years ago

3 0

Answer:

See below.

Step-by-step explanation:

Try each ordered pair in the equation. Each ordered pair is of the form (x, y). Replace x and y in the equation by values of x and y, respectively, in each ordered pair. Whichever ordered pair makes the equation a true statement is the answer.

For example:

Try (2, 3):

2x + 3y = 10

2(2) + 3(3) = 10

4 + 9 = 10

13 = 10

Since 13 = 10 is a false statement, (2, 3) is not a solution.

Try (2, 2):

2x + 3y = 10

2(2) + 3(2) = 10

4 + 6 = 10

10 = 10

Since 10 = 10 is a true statement, (2, 2) is a solution.

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What is the median of this data set? <br><br> Help!!!!!

max2010maxim [7]

Answer:

Step-by-step explanation:

4 0

3 years ago

Dr.Lamb is mixing a solution for his next patient. He has a cylindrical test tube which is 4 inches across on the inside. The de

SSSSS [86.1K]

Answer: There are 34.35 liters left in the test tube.

Step-by-step explanation:

Since we have given that

Depth of the tube = 9 inches

Circumference of base=4 inches

As we know the formula of circumference ,

$Circumference=2\pi r\\\\4=2\times \frac{22}{7}\times r\\\\4\times{7}{22\times 2}=r\\\\\frac{7}{11}=r$

Now, we need to find the volume of tube is given by

$\text{ Volume of test tube }=\pi r^2h\\\\\text{ Volume of test tube }=\frac{22}{7}\times \frac{7}{11}\times \frac{7}{11}\times 9\\\\\text{ Volume of test tube }=\frac{126}{11}=11.45\ inches^3$

Since we have also given that,

Of 1 cubic inch of space holds 3 liters of water,

So,

$1\text{ cubic inch}=3\text{ liters}\\\\11.45\text{ cubic inches}=3\times 11.45=34.35\text{ liters}$

Hence, there are 34.35 liters left in the test tube.

6 0

3 years ago

Can someone help me

Ulleksa [173]

Answer:

10°

Step-by-step explanation:

180-20 = 160

<BCA = 160°

AC=CB because its Radius

so the angle for <CBA and <BAD also the same

(180 - 160) ÷ 2 = <BAD

<BAD = 10°

4 0

3 years ago

Read 2 more answers

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :