All categories

3 years ago

Determine the common factor and HCF for each of the following terms.

Mathematics

1 answer:

Tamiku [17]3 years ago

8 0

Answer:

Step-by-step explanation:

1) 8y = 2 * 2 * 2 * y

12y = 2 * 2 * 3 * y

Common factors = 2 , 2 , y

HCF = 2*2y = 4y.

2) 2 b = 2 *b

3b = 3 * b

Common factor = b

HCF = b

3) 3w = 3 *w

5w² = 5 * w * w

Common factor = w

HCF = w

You might be interested in

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0

3 years ago

Jeremy's score was 1.75 standard deviations above the mean. which kf thr following is closest to his percentile rank

Gelneren [198K]

The answer is 96%.

Explanation:
It is generally presumed that the scores are normally distributed.

1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75

2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599

3) Multiply the area by 100 in order to find the percentile:
0.9599 × 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.

Hence, the answer is 96%.

4 0

3 years ago

Find the next three terms in the pattern 9,4,-1,-6,-11 then describe the pattern

Yanka [14]

You would be subtracting by 5 so the next three would be -16, -21, -26

6 0

3 years ago

The cost of 50 pounds of beans is $? Pls help!!

maksim [4K]

Answer:

69,78!!

Step-by-step explanation:

Espero te sirva

4 0

3 years ago

You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling

BARSIC [14]

Answer:

(a) 15 m

(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

d(2) = 25 - (5/2)·4 = 15 . . . . meters

(b) The relative velocity at 2.4 seconds is ...

d'(t) = -5t

d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

= 26 5/36 m/s = 94.1 km/h

_____

Comment on following distance

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0

3 years ago