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Usimov [2.4K]
3 years ago
6

Help I do not know how do you do this ?

Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0
m\angle LKJ=360-m\angle KJM-m\angle JML-m\angle MLK\\
m\angle LKJ=360-132-48-m\angle MLK\\
m\angle LKJ=180-m\angle MLK\\\\
m\angle MLK=180-m\angle KLN\\
m\angle MLK=180-150\\
m\angle MLK=30^{\circ}\\\\
m\angle LKJ=180-30\\
\boxed{m\angle LKJ=150^{\circ}}
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How much work must be done on a 1000-kg car to increase its speed from 1 m/s
yarga [219]

Answer:

w=f*d

f=ma

f=1000*10

f=10000

w=10000j

4 0
3 years ago
If n(A) = P and n(B)=q then n(AxB) is (a ) p (b ) q ( c ) p+q ( d ) pq<br><br>​
Butoxors [25]

By way of example, suppose <em>A</em> = {1, 2, 3} and <em>B</em> = {<em>a</em>, <em>b</em>, <em>c</em>}. Then the Cartesian product of <em>A</em> and <em>B</em> is

<em>A</em> × <em>B</em> = {{1, <em>a</em>}, {1, <em>b</em>}, {1, <em>c</em>}, {2, <em>a</em>}, {2, <em>b</em>}, {2, <em>c</em>}, {3, <em>a</em>}, {3, <em>b</em>}, {3, <em>c</em>}}

That is, each element in <em>A</em> gets a pairing with each element in <em>B</em>, and for each pairing you have <em>n(A)</em> choices for the first element and <em>n(B)</em> choices for the second element.

So if <em>n(A)</em> = <em>p</em> and <em>n(B)</em> = <em>q</em>, then <em>n(A</em> × <em>B)</em> = <em>pq</em>.

6 0
3 years ago
Does anyone know this
Dmitriy789 [7]

Answer: you minus the equations you have to get the equation that you want so it all can add up to that *

Step-by-step explanation:

3 0
3 years ago
Wanda placed $100 in a savings account that pays 2% simple interest per year. After 6 months Wanda made another deposit of $100.
STatiana [176]

Answer:

\$203.01

Step-by-step explanation:

we know that

The simple interest formula is equal to

A=P(1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

<em>After 6 months</em>

t=6/12=0.5\ years\\ P=\$100\\r=2\%=2/100=0.02

substitute in the formula above

A_1=100(1+0.02*0.5)

A_1=100(1.01)

A_1=\$101

<em>At the end of one year</em>

t=6/12=0.5\ years\\ P=A_1+\$100=\$101+\$100=\$201\\r=2\%=2/100=0.02

substitute in the formula above

A_2=201(1+0.02*0.5)

A_2=201(1.01)

A_2=\$203.01

8 0
3 years ago
cameron and hunter have a push up competition and hunter did 17 push ups hunter did 2 fewer push ups than cameron how many push
GarryVolchara [31]

Answer:

19

Step-by-step explanation:

4 0
3 years ago
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