Can you help me? I am confused this is a hard question for me

Answers

Answer 1

Answer:

Answer:

960, 160 for each face / base

Step-by-step explanation:

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Sarah is 9 years old her grandfather is 90 how many times old as Sarah

Answers

10 times older than her grandfather

3/8 of the cast in a musical have to sing what fraction of the cast does not have to sing

Answers

so 100% of the cast is 8/8 right? so you would subtract 8/8-3= 5/8

so, 5/8 of the cast does not have to sing

The answer is simply 5/8. 5/8 of the cast does not have to sing.

According to the Fundamental Theorem of Algebra, the graph of f(x) = x2 - 4x + 3, has roots. From the graph we can see that it has zeros.

Answers

Answer:

The graph $f(x)=x^2-4x+3$ has two zeros namely 3 and 1.

Step-by-step explanation:

Consider the given equation of graph $f(x)=x^2-4x+3$

According to the Fundamental Theorem of Algebra

For a given polynomial of degree n can have a maximum of n roots.

Thus, for the given equation $f(x)=x^2-4x+3$ the degree of polynomial is 2 , thus the function can have maximum of 2 roots.

We know at roots the value of function is 0 that is f(x) = 0,

Substitute f(x) = 0 , we get, $f(x)=x^2-4x+3=0$

This is a quadratic equation, $x^2-4x+3=0$

We first solve it manually and then check by plotting graph.

Quadratic equation can be solved using middle term splitting method,

here, -4x can be written as -x-3x,

$x^2-4x+3=0 \Rightarrow x^2-x-3x+3=0$

$\Rightarrow x(x-1)-3(x-1)=0$

$\Rightarrow (x-3)(x-1)=0$

Using zero product property, $a\cdot b=0 \Rightarrow a=0\ or \ b=0$

$\Rightarrow (x-3)=0$ or $\Rightarrow (x-1)=0$

$\Rightarrow x=3$ or $\Rightarrow x=1$

Thus, the two zero of f(x) are 3 and 1.

We can also see on graph attached below that the graph $f(x)=x^2-4x+3$ has two zeros namely 3 and 1.

Answer:

2 roots

2 zeros

Step-by-step explanation:

Two friends, Hailey and Jenna bought two tickets to see a play at a local theater . Each ticket cost $17.25. Each friend donated $1 to the theater's youth program. The friends paid a total of $40.50 which it also includes convenience . How much did each friend paid for the convenience fee?

Answers

If you let the convenience fee equal x, then...

(2×17.25)+(2×1)+(2×x)= 40.50

There are 2 girls so we multiply all the charges by 2. Add them together and you'll get 40.50. Now if we solve for x, we will get the amount of the convenience fee.

2×17.25 + 2×1 + 2x = 40.50
34.50 + 2 + 2x = 40.50
36.50 + 2x = 40.50
2x = 40.50 - 36.50
2x = 4
x = 2

Each girl paid a 2 dollar convenience fee.

Isabella spends $2.25 on poster boards. How many poster boards does she buy. Posterboards are $0.75.

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She can buy 3 because $0.75 times 3 is equal to $2.25

if u multiply.75 3 times it will = 2.25 or multiply 75×3=225

Solve irrational equation pls

Answers

$\hbox{Domain:}\nx^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\nx^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\nx(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\n(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\nx\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\nx\in(-\infty,-2\rangle\cup\langle3,\infty)$

$√(x^2+x-2)+√(x^2-4x+3)=√(x^2-1)\nx^2-1=x^2+x-2+2√((x^2+x-2)(x^2-4x+3))+x^2-4x+3\n2√((x^2+x-2)(x^2-4x+3))=-x^2+3x-2\n√((x^2+x-2)(x^2-4x+3))=(-x^2+3x-2)/(2)\n(x^2+x-2)(x^2-4x+3)=\left((-x^2+3x-2)/(2)\right)^2\n(x+2)(x-1)(x-3)(x-1)=\left((-x^2+x+2x-2)/(2)\right)^2\n(x+2)(x-3)(x-1)^2=\left((-x(x-1)+2(x-1))/(2)\right)^2\n(x+2)(x-3)(x-1)^2=\left((-(x-2)(x-1))/(2)\right)^2\n(x+2)(x-3)(x-1)^2=((x-2)^2(x-1)^2)/(4)\n4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\n$
$4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\n(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\n(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\n(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\n(x-1)^2(3x^2-28)=0\nx-1=0 \vee 3x^2-28=0\nx=1 \vee 3x^2=28\nx=1 \vee x^2=(28)/(3)\nx=1 \vee x=\sqrt{(28)/(3)} \vee x=-\sqrt{(28)/(3)}\n$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\nx\in\langle1,2\rangle\n$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{(28)/(3)}, -\sqrt{(28)/(3)}\}\n\boxed{\boxed{x=1}}$

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