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2 years ago

8

Heather has a set of 10 numbers, with no two the same. She adds the number 44 to the set, and the set has no mode. Which stateme

nt MUST be true about her original set

1 answer:

dsp732 years ago

8 0

Answer:

The number 44 was not in the original set.

Step-by-step explanation:

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One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

or

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

In JKL, k=4.1 cm,j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .

spayn [35]

Answer:

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

$\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

$\frac{k}{sin(K)}=\frac{j}{sin(J)} \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)} \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o \\$

6 0

2 years ago

A bus went halfway from a village to a city at a certain speed. Then it stopped in traffic for an hour. In order to catch up the

k0ka [10]

We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.

Since 200 km is "halfway", the total distance must be 400 km.

time = distance / speed
total time = (time for first half) + (delay) + (time for second half)

400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40

The speed of the bus before the traffic holdup was 40 km/h.

4 0

2 years ago

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2 years ago

Enter the coordinates of the solution of these two linear equations.

makvit [3.9K]

Answer:

(-4,2) are solution coordinates

This is the point in which these two lines intersect

6 0

2 years ago