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Vika [28.1K]
3 years ago
11

Which would be used to solve this equation? Check all that apply.

Mathematics
1 answer:
barxatty [35]3 years ago
7 0
Whats the equation ?
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What are the operations used to <br> solve this equation5/6x-2=-2/3x+1
SashulF [63]
Addition
multiplication
4 0
3 years ago
454 / 762 = ?
Maslowich

Answer:

huh-

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
g A shopping center includes a grocery store and a drug store. One%E2%80%8B afternoon, a total of 200 people visited the shoppin
QveST [7]

Answer:

a) 179 people shopped at the grocery store or the drug store.

b) 21 people shopped at neither the grocery store nor the drug store

Step-by-step explanation:

I am going to treat these events as Venn sets.

I am going to say that:

Set A: Shopped at the grocery store

Set B: Shoped at the drug store.

121 shopped at the grocery store:

This means that A = 121

91 shopped at the drug store:

This means that B = 91

33 shopped at both businesses.

This means that A \cap B = 33

a) How many people shopped at the grocery store or the drug store?

This is

A \cup B = A + B - (A \cap B)

With the values given in the exercise.

A \cup B = A + B - (A \cap B) = 121 + 91 - 33 = 179

179 people shopped at the grocery store or the drug store.

b) How many people shopped at neither the grocery store nor the drug store?

179 of 200 shopped in at least one. So 200 - 179 = 21 shopped at neither.

21 people shopped at neither the grocery store nor the drug store

8 0
2 years ago
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Vesnalui [34]
Since they are independent events to find the probability of both is P(A) * P(B)

P(A) = P(Heads) = \frac{1}{2}
P(B) = P(Roll ≥ 4) = \frac{3}{6} =  \frac{1}{2}

Now multiply those fractions together

\frac{1}{2} *  \frac{1}{2}  =  \frac{1}{4} = P(Heads & ≥ 4)
5 0
3 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
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