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serg [7]
3 years ago
9

GUYS HELP!!!

Mathematics
2 answers:
Maurinko [17]3 years ago
7 0
There isn’t enough information for us to help you:(
ch4aika [34]3 years ago
7 0

Answer:

hey we need a picture of the angles otherwise we cant help u sorry

You might be interested in
3.14*16.8*16.8*16.9
vitfil [10]

Answer: 3.14 *16.8*16.8*16.9=14977.3478

14977.3478= 14,977.35

6 0
3 years ago
If Julia has 20 cupcakes, and 10 people eat one cupcake, how much cupcakes are left?
Feliz [49]
There are going to be 10 cupcakes left,hope I helped (:
8 0
3 years ago
Read 2 more answers
Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)
DiKsa [7]

Answer:

The standard equation of the sphere is (x-5)^{2} + (y-5)^{2} + (z-4)^{2}  = 53

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = (\frac{x_{1} + x_{2}  }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2}  }{2})

Let the first endpoint be represented as (x_{1}, y_{1}, z_{1}) and the second endpoint be (x_{2}, y_{2}, z_{2}).

Hence,

Midpoint = (\frac{x_{1} + x_{2}  }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2}  }{2})

Midpoint = (\frac{3 + 7  }{2}, \frac{-2+12 }{2}, \frac{4 + 4  }{2})

Midpoint = (\frac{10 }{2}, \frac{10}{2}, \frac{8  }{2})\\

Midpoint = (5, 5, 4)

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2}      }

d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}

d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}

d = \sqrt{16 +196 + 0

d =\sqrt{212}

d = 2\sqrt{53}

This is the diameter

To find the radius, r

From Radius = \frac{Diameter}{2}

Radius = \frac{2\sqrt{53} }{2}

∴ Radius = \sqrt{53}

r = \sqrt{53}

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is (5, 5, 4)

Radius of the sphere is \sqrt{53}

The equation of a sphere of radius r and center (h,k,l) is given by

(x-h)^{2} + (y-k)^{2} + (z-l)^{2}  = r^{2}

Hence, the equation of the sphere of radius \sqrt{53} and center (5, 5, 4) is

(x-5)^{2} + (y-5)^{2} + (z-4)^{2}  = \sqrt{(53} )^{2}

(x-5)^{2} + (y-5)^{2} + (z-4)^{2}  = 53

This is the standard equation of the sphere

6 0
3 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
The length of a photograph is 1 cm less than twice the width. The area is 91cm^2. find the width and length.
maksim [4K]

Answer:

Width=6.5 cm

Length=12 cm

Step-by-step explanation:

Step 1: Express the lengths and widths

Width=w

Length=l, but 1 cm less than twice the width=(2×w)-1=2 w-1

Step 2: Solve for the length and width

A=L×W

where;

A=area of the photograph

L=length of the photograph

W=width of the photograph

In our case;

A=91 cm²

L=2 w-1

W=w

91=(2 w-1)w

2 w²-w=91

2 w²-w-91=0, is a quadratic equation

solve for w

w={-1±√(-1²-4×2×-91)}/(2×2)

w=(-1±27)/4

w=(27-1)/4=6.5, or (-1-27)/4=-8

Take w=6.5 cm

L=(2×6.5)-1=13-1=12 cm

Width=6.5 cm

Length=12 cm

6 0
3 years ago
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