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2 years ago

Can you please help me with this problem

Mathematics

1 answer:

Aleks04 [339]2 years ago

8 0

PASS IS INTERCEPTED AT THE GOAL LINE BY MALCOLM BUTLER

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Which triangle has 0 reflectional symmetries?

Tju [1.3M]

Answer:

the traingle that have 0 reflectional is the scalene traingle then its number one all defferent sides

give me brainliest if it true

3 0

2 years ago

Read 2 more answers

Pls, help anyone- it's due tomorrow!

harina [27]

Answer: 65.94

Step-by-step explanation:

Find the area of the middle circle.

3.14 x 2² = 12.56

Pretend the white area doesn't exist and find the area of the shaded region.

3.14 x 5² = 78.5

Then subtract.

78.5 - 12.56 = 65.94

6 0

2 years ago

X²-x-42

oee [108]

Answer:

The least common multiple (LCM) of two or more non-zero whole numbers is the smallest whole number that is divisible by each of those numbers. In other words, the LCM is the smallest number that all of the numbers divide into evenly. ( x − 7 ) ( x + 6 ) ( x + 7 ) ( x 2+ 66 ) ( x − 4 )

Step-by-step explanation:

4 0

2 years ago

What is the slope of the line that contains these points?

ryzh [129]

Answer:

-8

Step-by-step explanation:

Apply slope/gradient formula: (12 - 36)/(11 - 8) = -24/3 = -8

This is a question on coordinate geometry. If you wish to venture further into it/understand this topic better, you may want to follow my Instagram account (learntionary), I have already posted the slope formula which you may want to save it for future reference and I'll also post some of my own notes on certain topics and also some tips that may be useful to you :)

6 0

2 years ago

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

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

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago

Can you please help me with this problem ​

Can you please help me with this problem