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3 years ago

Can someone help me out please??

Mathematics

1 answer:

olganol [36]3 years ago

8 0

Answer:

the 2nd one is the answer.

x ≥ −8

Step-by-step explanation:

Hope that helps

Sorry if wrong

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Which equation represents this problem? Carol's age is three more than two times Lindsay's age. Carol is 39 years old. How old i

dem82 [27]

Answer:

2l + 3 = 39

Step-by-step explanation:

Let l = Lindsay's age in years.

Carol's age is three more than two times Lindsay's age

Carol = 2l+3

Carol = 39

39 = 2l+3

3 0

3 years ago

Read 2 more answers

Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0

4 years ago

Determine the coordinator of the point of intersection of lines 3x-2y=13 and 2y+x+1=0

finlep [7]

Answer:

(3,-2)

Step-by-step explanation:

Given equations of line

3x-2y=13

2y+x+1=0

=> x = -1 -2y

Point of intersection will coordinates where both equation have same value of (x,y)

top get that we have to solve the both equations by using method of substitution of simultaneous equation.

using this value of x in 3x-2y=13, we have

3(-1-2y) -2y = 13

=> -3 -6y-2y = 13

=> -8y = 13+3 = 16

=> y = 16/-8 = -2

x = -1 - 2y = -1 -2(-2) = -1+4= 3

Thus, point of intersection of line is (3,-2)

3 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago

10 + 9 • (−3) 2− (−1 + 8)

Vitek1552 [10]

Answer:

-51

Step-by-step explanation:

PEMDAS suggests you start with parenthesis

10+9•(-3)2-(7)

10-(27)2-(7)

10-54-7

-51

3 0

3 years ago

Read 2 more answers

Can someone help me out please??​

Can someone help me out please??