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3 years ago

Which figures can have the same cross section as a sphere?

Mathematics

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

Step-by-step explanation:

the cross section of a sphere is a circle

only the cone can have a circle for a cross section

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I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

professor190 [17]

Answer:

Step-by-step explanation:

The formula for the Pythagorean Theorem is:

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

Now,

A=7

B=24

7x7=49

24x24=576

Add,

576+49=625

Last,

Take the square root of 625

The square root of 625 is 25

c=25

To Check if the hypotenuse is correct:

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

$7^2+24^2=25^2$

7x7=49

24x24=576

25x25=625

49+576=625

625=625,

Therefore the hypotenuse is 25.

5 0

2 years ago

What value of x makes the equation -2/3x + 3 1/3x - 1/2 = -1/3x + 5 1/2 true? Please help ASAP!!!!!!!!

Naddika [18.5K]

Answer:

Step-by-step explanation:

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The steps of the solution in the photo above

I hope that is useful for you :)

3 0

2 years ago

What is T in -5t-2(5t+10)=100

Alina [70]

Answer:

t=-8

Step-by-step explanation:

-5t-2(5t+10t)=100

-2 x 5t=-10t

-2 x 10 = -20

-5t-10t-20=100

combine terms and add 20

-15t=120

divide by -15

t=-8

7 0

3 years ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago

4200 and 2100 simplified to there lowest terms

olchik [2.2K]

4000 for 4200
2000 for 2100

(If you meant a different answer please explain)

5 0

3 years ago

Read 2 more answers