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3 years ago

PLEASE HELP, TIMED

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Answer:

your answer is c hope this helps let me know if im wrong or nah

Step-by-step explanation:

OLga [1]3 years ago

5 0

I'm pretty sure your answer is C .

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Scientific notation 0.0094592 Scientific notation 0.94592

Elenna [48]

Answer:

9.4592x $10^{-3}$

94592x10^-1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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

2 years ago

Identify the pattern then write the next three terms in each sequence 192,96,48,24,...

Karo-lina-s [1.5K]

1st. Find the pattern
Pattern is dividing by 2.
24/2=12
12/2=6
6/2=3

4 0

3 years ago

A metalworker has a metal alloy that is 15% copper and another alloy that is 60% copper. How many kilograms of each alloy shou

coldgirl [10]

18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.

Step-by-step explanation:

Let x = kg of 15% copper alloy

Let y = kg of 60% copper alloy

Since we need to create 90 kg of alloy we know:

x + y = 90

51% of 90 kg = 45.9 kg of copper

So we're interested in creating 45.9 kg of copper

We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:

0.15x + 0.60y = 45.9

but

x + y = 90

x= 90 - y

substituting that value in for x

0.15(90 - y) + 0.60y = 45.9

13.5 - 0.15y + 0.60y = 45.9

0.45y = 32.4

y = 72

Substituting this y value to solve for x gives:

x + y = 90

x= 90-72

x=18

Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.

6 0

3 years ago

1. What is the last step in constructing an angle? Help please For 30 points!!!

zaharov [31]

Answer:

The second choice. Connect the endpoint and the mark at the angle measure.

Step-by-step explanation:

4 0

3 years ago