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2 years ago

8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?

Mathematics

1 answer:

Lerok [7]2 years ago

8 0

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

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A right triangle has legs of length 4x and 20x. What is an expression for the hypotenuse of the right triangle

kherson [118]

Answer:

$4x\sqrt{26}$ .

Step-by-step explanation:

Given information:

$Leg_1=4x$

$Leg_2=20x$

According to the Pythagoras theorem,

$(hypotenuse)^2=(Leg_1)^2+(leg_2)^2$

Using Pythagoras theorem, we get

$(hypotenuse)^2=(4x)^2+(20x)^2$

$(hypotenuse)^2=16x^2+400x^2$

$(hypotenuse)^2=416x^2$

Taking square root on both sides.

$hypotenuse=\sqrt{416x^2}$

$hypotenuse=4x\sqrt{26}$

Hence, the expression of the hypotenuse is $4x\sqrt{26}$ .

5 0

2 years ago

Which best describes the graph of the cubic function f(x) = x3 + x2 + x + 1?

EleoNora [17]

Answer:

Option A

Step-by-step explanation:

f(x) = x³ + x² + x + 1

This is increasing function

The first option is correct

<em>See attached graph for reference</em>

3 0

2 years ago

A remote-controlled plane is flying 16.5 meters above the ground. The plane dives 8 meters

s2008m [1.1K]

Answer:

Step-by-step explanation:

3 0

2 years ago

At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima

Firdavs [7]

$\mathbf r(t)=\langle x(t),y(t),z(t)\rangle=\langle\sin t,\cos t,t\rangle$

$x^2+y^2+z^2=\sin^2t+\cos^2t+t^2=65$
$\implies t^2=64$
$\implies t=\pm8$

7 0

2 years ago

A row of tiny red beads is placed at the beginning and at the end of a 20‐centimeter bookmark.

AlladinOne [14]

Answer:

51 rows

Step-by-step explanation:

Given

Length of bookmark = 20cm

Distance between beads = 4mm

Required

Number of rows of beads

First, the distance between the rows of beads must be converted to cm

if 1mm = 0.1cm

then

4mm = 4*0.1cm

4mm = 0.4 cm

This means that each row of beads is placed at 0.4 cm mark.

The distance between each row follows an arithmetic progression and it can be solved as follows;

$T_n = a + (n-1)d$

Where $Tn = 20cm$ (The last term)

$a = 0 cm$ (The first term)

$d = 0.4cm$ (The distance between each row of beads)

n = ?? (number of rows)

Solving for n; we have the following;

$T_n = a + (n-1)d$ becomes

$20 = 0 + (n-1)0.4$

$20 = (n-1)0.4$

DIvide both sides by 0.4

$\frac{20}{0.4} = \frac{(n-1)0.4}{0.4}$

$50 = (n-1)$

$50 = n-1$

Add 1 to both sides

$50 + 1= n-1 + 1$

$n = 51$

Hence, the number of rows of beads is 51

4 0

2 years ago