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attashe74 [19]
3 years ago
9

I am having trouble with my homeschool homework. please help.

Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0
Hello,
half of 3/4 =1/2*3/4=3/8

3/8+1/12+/18=3/8+1/8+1/12=1/2+1/12=7/12

Answer D
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What is 0.506 rounded in the hundredths
Aleksandr [31]
0.51 The hundredths place increases when it is higher than or equal to 5 in the thousandths place.

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If x+y = z and y−z=x which of the following must be equal to z?
umka21 [38]

<u>ANSWER</u>

(C) y

<u>EXPLANATION</u>

We have two equations:

The first one is

x + y = z...(1)

The second equation is

y - z = x...(2)

Let us make z the subject of this second equation:

\implies \: -x +y = z...(3)

We now add equation (1) and (3) to get:

(-x + x) + (y +  y) = z + z

We simplify to get:

0+2y= 2z

2y= 2z

Divide throu by 2.

\frac{2y}{2}  =  \frac{2z}{2}

y= z

\therefore \: z = y

The correct answer is C

8 0
4 years ago
5. x2 – 10x + 21 = 0
andrezito [222]
TRY PHOTOMATH it’ll help u with questions like this , all u do is take a pic and it’ll show u the answer and how to do it
7 0
3 years ago
The ratio of tagged fish to untagged fish was 2 to 9. Ninety fish were tagged. How many fish were untagged
Nimfa-mama [501]

Answer:

405

Step-by-step explanation:

Since

2 : 9 = 90 : X

Then we know

9/2 = X/90

Multiplying both sides by 90 cancels on the right

90 × (9/2) = (X/90) × 90

90 × (9/2) = X

Then solving for X

X = 90 × (9/2)

X = 405

Therefore

2 : 9 = 90 : 405

5 0
3 years ago
Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(
balu736 [363]

It looks like we're told that

T(-2x^2)=3x^2+4x

T\left(\dfrac12x+3\right)=-3x^2+3x-4

T(2x^2-1)=-3x+4

We use the fact that T is linear to find T(1),T(x),T(x^2). First, we notice that

T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)

We also have

T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)

T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)

So once we find T(1), we can determine T(x) and T(x^2). We have

T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4

and using this we find

T(x)=12x^2+12x+16

T(x^2)=-\dfrac32x^2-2x

Then

T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)

T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c

8 0
4 years ago
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