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3 years ago

90 out of 120 students in your grade are girls at your school. What percentage are boys?

Mathematics

1 answer:

sesenic [268]3 years ago

7 0

30% Bc there are 30 boys and 90 girls or it’s 10% boys

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Instead of sharing with friends, Kaleb decides to share his bag of LEGO bricks with his sister, Kami. Kaleb randomly selects one

Furkat [3]

The black brick chance would be 45/117

3 0

3 years ago

A ski lodge charges $26 per day for a pass and $1.50 for every ski lift ride. A second ski

malfutka [58]

Answer:

32=40

26=39

Step-by-step explanation:

8 0

2 years ago

(5 hundreds 5 tens)×10 in unit form what's the unit form?

Neko [114]

The unit form for that is
5,500

5 0

2 years ago

The equation for line r can be written as y=-1/4x+3. Line s which is parallel to lone r includes the point (-4,3). What is the e

MaRussiya [10]

Answer:

$y=-\frac{1}{4}x+2$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when x is 0)
Parallel lines always have the same slope

1) Determine the slope of line S using line R (m)

$y=-\frac{1}{4} x+3$

We can identify clearly that the slope of the line is $-\frac{1}{4}$ , as it is in the place of m. Because parallel lines always have the same slope, the slope of line S would also be $-\frac{1}{4}$ . Plug this into $y=mx+b$ :

$y=-\frac{1}{4}x+b$

2) Determine the y-intercept of line S (b)

$y=-\frac{1}{4}x+b$

Plug in the given point (-4,3) and solve for b

$3=-\frac{1}{4}(-4)+b\\3=1+b$

Subtract 1 from both sides to isolate b

$3-1=1+b-1\\2=b$

Therefore, the y-intercept is 2. Plug this back into $y=-\frac{1}{4}x+b$ :

$y=-\frac{1}{4}x+2$

I hope this helps!

6 0

2 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