All categories

3 years ago

Which improper fraction shows the value of the shaded portion of the image?

Mathematics

1 answer:

Debora [2.8K]3 years ago

3 0

Answer:

i dont know

Step-by-step explanation:

You might be interested in

What is the range of ƒ(x) = c?

nikdorinn [45]

All real numbers because it’s a function that has no limit

3 0

2 years ago

AC = 16, AB = x + 1, and BC = x + 7. What is the measure of the length of AB? HELP

saveliy_v [14]

Answer:

Step-by-step explanation:

AC=AB+BC

so 16=x+1+x+7

which simplifies to 16=2x+8

subtract eight from both sides to get 8=2x

then divide by 2 to get that x=4

AB=x+1, which substitutes into 4+1=5

6 0

2 years ago

Thomas reads 2/9 of a book on Monday and 1/6 of it on Tuesday. He reads twice as many pages on Wednesday as on Tuesday. What fra

dezoksy [38]

1/18 is left. Thomas has 1/18 left of his book. How? What denominator can 9&6 relate to? 54? Sure, but it could be lower. How about 18? Ok. We have 18 because 9x2 and 6x3 both equal 18... 2/9 becomes 4/18 because 18 divided by 9 is 2. You multiply 2 with the two from 2/9. 1/6 becomes 3/18 because 18 divided by 6 is 3. You then multiply 3 with the one from 1/6. Put it together and TA-DA... Your answer is 1/18

8 0

3 years ago

Calculus 3 chapter 16

o-na [289]

Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$

3 0

1 year ago

Complete the proof by providing the missing reasons. given:cb bisects bd,de bisects ec, cb=de. prove dbc=ced. . . CB=DE,CB BISEC

Alexus [3.1K]

Base on your question were the are proofs that is given are this are: CB bisects BD, DE bisects EC, CB=DE. So answer your question you must understand first your given statements and realize that CBD and DEC are both right triangles because they are a perpendicular segment so i conclude that the answer should be CBD=DEC because they are both right angles

5 0

2 years ago

Read 2 more answers