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vampirchik [111]

4 years ago

10

Lodi solves the equation 3+ x/2=10 and justified each step as shown. Identify Lodi’s error and fix his mistake. Remember the jus

tification is Lodi’s work as well.

1 answer:

ad-work [718]4 years ago

8 0

3+ x/2 = 10 Given

x/2 = 7 Subtraction property of equality

x = 14 Multiplication property of equality

You might be interested in

Write a quadratic function in standard form with zeros 1 and -10

Licemer1 [7]

Answer:

f(x)=x^2+9x-10

Step-by-step explanation:

<u>Standard Form of Quadratic Function</u>

The standard form of a quadratic function is:

$f(x)=ax^2+bx+c$

where a,b, and c are constants.

The factored form of a quadratic equation is:

$f(x)=a(x-\alpha)(x-\beta)$

Where $\alpha$ and $\beta$ are the roots or zeros of f, and a is constant.

We know the zeros of the function are 1 and -10. The function is:

$f(x)=a(x-1)(x-(-10))$

$f(x)=a(x-1)(x+10)$

Operating:

$f(x)=a(x^2+10x-x-10)$

Joining like terms:

$f(x)=a(x^2+9x-10)$

Since we are not given any more restrictions, we can choose the value of a=1, thus. the required function is:

$\boxed{f(x)=x^2+9x-10}$

6 0

3 years ago

Derek sees that a 24 foot flagpole casts a 10 foot shadow. At the same time of day, if a nearby tree casts

Lunna [17]

Answer:

60 ft.

Step-by-step explanation:

We can solve this problem with a simple proportion:

24 / 10 = h / 25

10h = 24 * 25

600 = 10h

60 ft. = h

4 0

3 years ago

Please help me on this I don't know how to do this help me to find out what property it is thank you . .

Agata [3.3K]

Distributive
it has parenthese which you have to multiply with
x= infinite solution btw

8 0

3 years ago

Which geometric figures are drawn on the diagram?<br> Check all that apply.

Allisa [31]

Try adding a picture next time maybe...?

7 0

3 years ago

Read 2 more answers

Please help me thank you

Hoochie [10]

Answer:

$\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}$

Step-by-step explanation:

We know:

$\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet$

We have

$\sin\theta=\dfrac{1}{2}$

Use $\sin^2\theta+\cos^2\theta=1$

$\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}$

$\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}$

3 0

3 years ago