Makesha lost 21 pounds in 12 weeks. Find her rate loss in pounds per week.

Triangle ABC is an isosceles triangle. Angles B and C are base angles, with measurements of (11x −10)° and (7x + 10)°, respectiv

Ulleksa [173]

Answer:

m∠A = 90°

Step-by-step explanation:

In isosceles triangle base angles are congruent. That means we can equate measurments of angle B and angle C and solve for x!

m∠B = m∠C

11x - 10 = 7x + 10

4x - 10 = 10

4x = 20

x = 5

Now let's insert x back in the expressions for angles.

m∠B = (11x − 10)° = (11(5) − 10)° = 45°

m∠C = (7x + 10)° = (7(5) + 10)° = 45°

<u>Sum of all angles in the triangle is 180°.</u> Let's make an equation.

m∠A + m∠B + m∠C = 180°

m∠A + 45°+ 45° = 180°

m∠A = 90°

7 0

1 year ago

Please Help Fast!! 15 Point and brainiest if correct

iragen [17]

If a < b, then dividing both sides by a positive number will not flip the inequality sign. The inequality flips if you divide both sides by a negative number.

So if a < b, then a/3 < b/3 as well.

<h3>Answer: Less than sign (first choice)</h3>

6 0

3 years ago

Mr. Lopez wrote the equation 32g+8g-10g=150 on the board. Four students explained how to solve for g. Alyssa: “I added 32 and 8

Vaselesa [24]

Answer:

Wilhem

Step-by-step explanation:

The given equation is

$32g+8g-10g=150$

Step 1 : Add 32g and 8g.

$(32g+8g)-10g=150$

$(32+8)g-10g=150$

$40g-10g=150$

Step 2: Subtract 10 from 40.

$(40-10)g=150$

$30g=150$

Step 3: Divide both sides by 30, find the value of g.

$\frac{30g}{30}=\frac{150}{30}$

$g=5$

The value of g is 5.

Therefore, Wilhem is correct.

7 0

2 years ago

Read 2 more answers

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: