All categories

3 years ago

Please help with this problem!

Mathematics

1 answer:

Viktor [21]3 years ago

6 0

Answer:

Second option

Step-by-step explanation:

Let's assume that x is 1 and y is 3

a) $\frac{1}{2}$ < $\frac{3}{2}$ = true

b) -1 < -3 = false

c) 1+2 < 3+2 = true

d) 2 x 1 < 2 x 3 = true

You might be interested in

3/7 x =5/14<br> solve for X

Paul [167]

Answer:

$\frac{5}{6} = x$

Step-by-step explanation:

$\frac{3}{7}x = \frac{5}{14} \\ \\ \frac{5}{14} \div \frac{3}{7} = \frac{5}{14} \times \frac{7}{3} = \frac{35}{42} = \frac{5}{6}$

Use multiplication [inverse of division] to figure this out.

* 2⅓ = 7⁄3

I am joyous to assist you anytime.

7 0

3 years ago

Help me with this. will mark as brainliest

Pani-rosa [81]

Half of R is shown as 20 degrees, so the other half of r is also 20 degrees.

A tangent line forms a right angle, so the tangent at P is 90 degrees.

X would be 180 - 90 - 20 = 70 degrees.

X = 70 degrees.

8 0

3 years ago

If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,

S_A_V [24]

Taking $y=y(x)$ and differentiating both sides with respect to $x$ yields

$\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

Solving for the first derivative, we have

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y$

Differentiating again gives

$\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0$

Solving for the second derivative, we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}$

Now, when $x=1$ and $y=2$ , we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63$

3 0

3 years ago

Gunther invested $10,000 in two mutual funds. One of the funds rose 6% in one year and the other rose 9% in one year. If his inv

mote1985 [20]

Answer:

$7200 in the fund rose 6% n $2800 in the fund rose 9%

Step-by-step explanation:

let x be the amount in the fund rose 6%

gain in one year=x*6%=0.06x

total amount is 10000

so amount in the fund rose 9% = 10000-x

gain in one year=(10000-x)*9%=900-0.09x

total gain=0.06x+900-0.09x=900-0.03x

=684

900-684=0.03x

0.03x=216

x=7200

the other fund amount=10000-7200=2800

5 0

3 years ago

Read 2 more answers

How do I convert 22.4 kg/L to kg/mL

olya-2409 [2.1K]

You have to move the decimal

5 0

3 years ago