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Vikki [24]
3 years ago
10

Thirty students are at soccer practice. Each group of three students needs a soccer ball and two field cones . How many balls an

d cones are needed in all? balls ___. cones ____.
Mathematics
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

10 soccer balls

20 cones

Step-by-step explanation:

30/3=10 groups

10*2=20

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8x-5y= 14<br> – 5х+бу = -3
podryga [215]

Answer:

y= 34/23

x=  123/46

Step-by-step explanation:

Solve for x

8x-5y= 14

x= 14/8 +5/8y

Sub x into second equation

– 5х+бу = -3

-5(14/8 +5/8y) + 6y= -3

-29/4 +23/8y = -3

y= 34/23

Sub y into any equation and solve for x

8x=483/23

x= 123/46

4 0
3 years ago
During the first year Sherry worked as a lifeguard, she was paid at a rate of $12.75 per hour.
lara [203]

Answer:

a

Step-by-step explanation:

5 0
3 years ago
Find the slope of the line passing through the points (2,-9) and )1,-3)
nignag [31]

Hi there!

\large\boxed{\text{slope =} 6}

We can calculate slope using the following formula:

slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Plug in the corresponding points:

slope = \frac{-9-(-3)}{2-1}

Simplify:

slope = \frac{-6}{1} = -6

3 0
2 years ago
Read 2 more answers
The circumference of a circle is 15 pie centimeters. What is the area of the circle in terms of pie
Jobisdone [24]

Answer:

C

Step-by-step explanation:

The circumference (C) of a circle = 2πr ← r is the radius

Given C = 15π the equating gives

2πr = 15π ( divide both sides by 2π )

r = \frac{15}{2} = 7.5 cm

The area (A) of a circle is calculated using the formula

A = πr² = π × 7.5² = 56.25π cm² → C

5 0
3 years ago
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Assume V and W are​ finite-dimensional vector spaces and T is a linear transformation from V to​ W, T: Upper V right arrow Upper
scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that  

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.  

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that  

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0  

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

8 0
2 years ago
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