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lana [24]
3 years ago
6

I really need help with this one it says 8 of the beads are purple in case it's hard to see

Mathematics
2 answers:
Harman [31]3 years ago
7 0
48% of the beads are purple
liraira [26]3 years ago
7 0
First of all, ask google before anything but im guessing the answer is 11 because I have 400k on insta and 90k on youtube subscribe! merch link in description
You might be interested in
Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
3 years ago
Marcus gardens. He pulls weeds for 18 minutes, Waters for 13 minutes, andplants for 16 minutes how many total minutes does he sp
Virty [35]

47 minutes gardening

7 0
3 years ago
Read 2 more answers
If 1once of toppings costs. .50 how much would 11onces be
dybincka [34]

Answer:

If one ounce of toppings cost $0.50, then you would need to take $0.50*11

.5*11= 5.5

11 ounces would be $5.50

hope this helps ;)

6 0
3 years ago
Please Help this is Due in 1 hour and I really dont understand any of this. (Question in Image)
vaieri [72.5K]

Answer:

(-4, -1/2)

Step-by-step explanation:

to calculate the midpoints bt of the line use the ormula:

( \frac{x1 + x2}{2} \frac{y1 + y2}{2} )

Where (x1, x2) is one coordinate point and (y1, y2) is anither coordinate point. Any two points will work, but I chose A (-5,-4) and B (-3, 3).

( \frac{ - 5 + ( - 3)}{2}  \frac{ - 4 + 3}{2} )

(-4, -1/2)

5 0
3 years ago
Help me please i would appreciate it alot (stop answering with things like 'im not sure' just put those in the comments)
OLEGan [10]
#3 would be positive
3 0
3 years ago
Read 2 more answers
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