Ask question

Ask question

All categories

2 years ago

9

Can you please help thanks

1 answer:

Svetach [21]2 years ago

4 0

I'll answer and write my work out -- I will post a link to the picture. It won't be in a diagram but it will tell you the answer. If you are on mobile you should log onto brainly on a computer so you can copy-paste the link into google

You might be interested in

17. What's the difference between 1,968 and 3,000? Check your answer using inverse operations.

Georgia [21]

The difference between 1,968 and 3,000 is 1,038. You can check the answer using inverse operations by adding 1,038 and 1,968. If it gives 3,000 then the answer is right. 3, 000 - 1,968 = 1,038.

1,038 + 1,968 = 3,000

6 0

2 years ago

What is the equation for twice a number gives 10

bearhunter [10]

Answer:5×2, at least i believe so, it could be asking for much more than that if I'm being honest

Step-by-step explanation:5×2=10

5 doubled is 10

5+5=10

4 0

2 years ago

Read 2 more answers

mount Rushmore is located 5,725 feet above sea level. About how high is Mount Rushmore above sea level, to the nearest thousand

DanielleElmas [232]

5,725 rounds to:

6,000 ft

7 0

2 years ago

What is the discriminant of the quadratic equation, 4x^2-8x-5=0

uysha [10]

Answer:

Step-by-step explanation:

4x^2-8x-5=0

The discriminant is the number under the √. In this case it's 144

7 0

2 years ago

Read 2 more answers

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

$y-y_1=m(x-x_1)$

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

5 0

2 years ago