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Aliun [14]
3 years ago
12

What is 1/4 times (3x7)

Mathematics
1 answer:
rjkz [21]3 years ago
4 0

Answer:

I believe it is 5.25

Step-by-step explanation:

if you multiply 3 and 7 then divide by 4 you will get 1 fourth of 21 which is 5.25 !!

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Write the expression in complete factored form <br><br> 3p^2(q+3)-4(q+3)
Tresset [83]
If you factored this the outcome would be,

(q+3)(3p^2-4)
6 0
3 years ago
Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y)(x,y) point.
Serga [27]

Answer:

(8,-3)

Step-by-step explanation:

The given parabola is

y =  - 3 {x}^{2}  + 48x  - 195

We factor -3 from the first two terms to get:

y =  - 3( {x}^{2}   - 16x)  - 195

Add and subtract the square of half the coefficient of x

y =  - 3( {x}^{2}   - 16x + 64)   + 3 \times 64- 195

y =  - 3 {(x - 8)}^{2}  + 192 - 195

We simplify to get:

y =  - 3 {(x - 8)}^{2}  - 3

We compare this to y=a(x-h)²+k,

The vertex is (h,k)=(8,-3)

3 0
3 years ago
Graph the function <br> g(x) = 2(x - 2)^2 + 2
hoa [83]

Answer:

g(x) = 2(x - 2)^2 + 2 plot (see attachment)

Step-by-step explanation:

8 0
3 years ago
Bill and George go target shooting together. Both shoot at atarget at the same time. Suppose Bill hits the target withprobabilit
german

Answer: (a) \frac{2}{9}       (b) \frac{6}{41}

Step-by-step explanation:

(a) P( Bill hitting the target) = 0.7        P( Bill not hitting the target) = 0.3

    P( George hitting the target) = 0.4     P(George not hitting the target) = 0.6

Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3

                                                                                            = 0.54

Chances that George hit the target is = 0.4 x 0.3 = 0.12

So given that exactly one shot hit the target, probability that it was George's shot = \frac{0.12}{0.54} = \frac{2}{9} .

(b) The numerator in the second part would be the same as of (a) part which is 0.12.

The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.

Given that the target is hit,probability that George hit it =                                                 \frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}  = = \frac{6}{41}                                                                        

                                                                                           

6 0
2 years ago
A:b = 1:5<br> a:c = 2:1<br> please helppp
joja [24]

Answer:

what is the question I don't understand the question so sorry

5 0
2 years ago
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