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3 years ago

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For a science project, 3/4 of the students chose to make a poster and 0.25 of the students wrote a report. Rosa said that more s

tudents made a poster than wrote a report. Do you agree with Rosa? Use numbers and words to explain your answer.

1 answer:

Viktor [21]3 years ago

3 0

She’s right because .25 converted into a fraction is 1/4 and 3/4 is greater than 1/4

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Write an equation in point-slope form for the line that passes through each point.

frutty [35]

Answer:

You can't make a point=slope form equation.

Step-by-step explanation:

The vertical line would be parallel to y- axis, hence its equation would be x= c . The slope of the vertical line is undefined, therefore there is no point slope form as such.

4 0

3 years ago

Read 2 more answers

What is the general form of the quadratic with vertex at (2, 3) and passing through the point (5, 12)?

ki77a [65]

The general form of the quadratic with vertex at (2, 3) and passing through the point (5, 12) is x²-4x+7.

<h3>What is the vertex form of a quadratic equation?</h3>

Vertex form of the quadratic equation is used to find the coordinate of vertex points at which it crosses its symmetry.

The standard equation of the vertex form of quadratic is given as,

$y=a(x-h)^2+k$

Here, (h, k) is the vertex point.

The vertex of quadratic at (2, 3) and passing through the point (5, 12). Put the value of vertex point in the above equation,

$y=a(x-2)^2+3$

As the quadratic passes through the point (5, 12). Thus, put the value of this point,

$12=a(5-2)^2+3\\12-3=a(3)^2\\9=a\times9\\a=\dfrac{9}{9}\\a=1$

Put the value of <em>a</em> in the above expression,

$y=1(x-2)^2+3\\y=(x-2)^2+3\\y=x^2-4x+4+3\\y=x^2-4x+7$

Thus, the general form of the quadratic with vertex at (2, 3) and passing through the point (5, 12) is x²-4x+7.

Learn more about the vertex form here;

brainly.com/question/17987697

#SPJ1

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3 years ago

The figure shows two parallel lines KL and NO cut by the transversals KO and LN.

Sever21 [200]

Answer:

(A)

Step-by-step explanation:

From the given figure, we have to prove whether the two given triangles are congruent or similar.

Thus, From the figure, ∠3=∠4 (Vertically opposite angles)

Since, KL and NO are parallel lines and KO and LN are transversals, then

measure angle 1= measure angle 5 that is ∠1=∠5(Alternate angles).

Thus, by AA similarity rule, ΔKLM is similar to ΔONM.

Thus, Option A that is Triangle KLM is similar to triangle ONM because measure of angle 3 equals measure of angle 4 and measure of angle 1 equals measure of angle 5 is correct.

8 0

3 years ago

Read 2 more answers

K=?<br><br> Can you also explain so I can try it by my self with a similar one.

Genrish500 [490]

Answer:

k = 10.39

Step-by-step explanation:

tangent 30 = opposite side / adjacent side, opposite side = 9, adjacent side = k + 3sqrt3.

tan 30 = 9/(k + 3sqrt3)

0.5774 = 9/(k + 3sqrt3)

k + 3sqrt3 = 15.5885

k = 15.5885 - 3sqrt3

k = 10.392

8 0

3 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

4 years ago