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3 years ago

For what value of b do the lines bx+3y=10 and x–2y=4 intersect at a point on the x-axis?

1 answer:

7 0

Since we need to find the point at which the 2 lines intersect at the x-axis, we need to find where the 2nd line intersects the x axis, by plugging in y=0.

x-2(0)=4
x=4
Therefore, at (4,0).

Now we plug in (4,0) at the first equation.
4b+3(0)=10
4b=10
b=5/2

Therefore, at b=5/2

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Trihao is driving from cambridge to london Assume the distance between cambridge and london is 100 km Given that trihao drives a

Paraphin [41]

average speed equals distance over time 100/80 will give you 5/4

8 0

1 year ago

18/34 as a reduced fraction

olya-2409 [2.1K]

I like to reduce everything by half if I don't already know the answer.

$\frac{18}{34} = \frac{9}{17}$

It just so happens that is the reduced fraction.

All I did was halve 18, which is 9, and halve 34, which is 17.

3 0

3 years ago

Express as a sum: x–y–z<br><br> plz help i will give branilly for the correct answer

babymother [125]

Answer:

i am positive it is x+y+z

Step-by-step explanation:

i'm so sorry if this is wrong..hope this helped :)

3 0

2 years ago

Identify the correct corresponding parts

andrey2020 [161]

Answer:

The correct corresponding part is;

$\overline {CB}$ ≅ $\overline {CD}$

Step-by-step explanation:

The information given symbolically in the diagram are;

ΔCAB is congruent to ΔCED (ΔCAB ≅ ΔCED)

Segment $\overline {CA}$ is congruent to $\overline {CE}$ ( $\overline {CA}$ ≅ $\overline {CE}$ )

Segment $\overline {CB}$ is congruent to $\overline {CD}$ ( $\overline {CB}$ ≅ $\overline {CD}$ )

From which, we have;

∠A ≅ ∠E by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∠B ≅ ∠D by CPCTC

Segment $\overline {AB}$ is congruent to $\overline {DE}$ ( $\overline {AB}$ ≅ $\overline {DE}$ ) by CPCTC

Segment $\overline {AE}$ bisects $\overline {BD}$

Segment $\overline {BD}$ bisects $\overline {AE}$

Therefore, the correct option is $\overline {CB}$ ≅ $\overline {CD}$

3 0

2 years ago

A truck weighing 150kN and travelling at 2m/sec impacts with a buffer

FrozenT [24]

The maximum compression of the spring is 0.276 m

Step-by-step explanation:

First of all, we calculate the spring constant of the spring, by using Hooke's law:

$F=kx$

where

F = 10 kN = 10,000 N is the force applied

x = 1.25 cm = 0.0125 m is the corresponding compression of the spring

k is the spring constant

Solving for k,

$k=\frac{F}{x}=\frac{10,000}{0.0125}=8\cdot 10^5 N/m$

When the truck impacts with the spring, all the kinetic energy of the truck is converted into elastic potential energy of the spring, so we can write

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

where

m is the mass of the truck

v = 2 m/s is the initial speed of the truck

$k=8\cdot 10^5 N/m$ is the spring constant

x is the compression of the spring

The mass of the truck can be found from its weight:

$m=\frac{W}{g}=\frac{150,000 N}{9.81 N/kg}=15,290 kg$

So now we can re-arrange eq.(1) to find the compression, x:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(15,290)(2)^2}{8\cdot 10^5}}=0.276 m$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

6 0

2 years ago