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3 years ago

DESPERATE PLS HELP

Mathematics

1 answer:

Luda [366]3 years ago

3 0

He will have $840 in 24 months, which is two years.

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It took the carriage 2.5 hours to travel 20 miles. H ow long will it take for the carriage to travel in 12 miles

Tomtit [17]

$\bf \begin{array}{ccll}
hours&miles\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
2.5&20\\
h&12
\end{array}\implies \cfrac{2.5}{h}=\cfrac{20}{12}\implies \cfrac{2.5\cdot 12}{20}=h$

6 0

3 years ago

If -2x+xy=30 and y=8,then what is the value of x?

Dafna1 [17]

-2x + xy = 30.....when y = 8

-2x + 8x = 30
6x = 30
x = 30/6
x = 5 <==

3 0

3 years ago

Read 2 more answers

What is the solution(s) to the system of equations y = x² +4 and y = 2x + 4

bija089 [108]

Answer:

$(x,y) = (0,4)~ \text{and}~ (x,y) = (2,8)$

Step by step explanation:

$y = x^2 +4~~~~~~~~~...(i)\\\\y = 2x +4~~~~~~~~~...(ii)\\\\\text{From (i) and (ii):}\\\\~~~~~~x^2 +4 = 2x+4\\\\\implies x^2 = 2x\\\\\implies x^2 -2x = 0\\\\\implies x(x-2) = 0\\\\\implies x =0,~ x = 2$

$\text{Substitute x = 0 in eq (i):}\\\\y = 0^2 +4 = 4\\\\\text{Substitute x = 2 in eq (i):}\\\\y=2^2+4 = 4+4 = 8\\\\\text{Hence,}~ (x,y)= \{(2,8), (0,4)\}$

3 0

2 years ago

Can I get help solving this graph please?

Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

8 0

3 years ago

A car is traveling at 60 mph and is tailgating another car at distance of only 30 ft. If the reaction time of the tailgater is 0

Sveta_85 [38]

Answer:

44 ft

Step-by-step explanation:

We are given that;

Speed; v = 60 mph

Reaction time;t = 0.5 s

Now, we know that;

Distance = speed x time

Since time is in seconds, let's convert speed to mph to ft/s

From conversion,

1 mph = 1.46667 ft/s

So, 60 mph would give;

60 × 1.46667 ≈ 88 ft/s

Thus, distance his car travels during his reaction time is;

Distance = 88 × 0.5 = 44 ft

6 0

3 years ago