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3 years ago

What is d in this problem d⋅24=−12

Mathematics

2 answers:

brilliants [131]3 years ago

8 0

its 8 lololololololololololololololol

kobusy [5.1K]3 years ago

4 0

The answer is 8

Jzbdbdndnbd

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If f(x)=81, what is x?

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There's not enough information...

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3 years ago

PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM

Brrunno [24]

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

$c^{2} =a^{2} +b^{2}$

c = $\sqrt{a^{2} +b^{2} }$

For question 1:

c = ?

a = 40

b = 9

putting them in formula,

c = $\sqrt{40^{2} + 9^{2} }$

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = $\sqrt{12^{2} + 13^{2} }$

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

$c^{2} =a^{2} +b^{2}$

$35^{2} = 20^{2} + b^{2}$

1225 = 400 + $b^{2}$

$b^{2}$ = 1225 - 400

$b^{2}$ = 825

$\sqrt{b^{2} } = \sqrt{825}$

b = 5 $\sqrt{33}$

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

$c^{2} =a^{2} +b^{2}$

$37^{2} = 20^{2} + b^{2}$

1369 = 400 + $b^{2}$

$b^{2}$ = 1369 - 400

$b^{2}$ = 969

Taking square root on both sides

b = 31.12

Hope it helps.

4 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