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2 years ago

Help me plz ill be dead if i dont do this

Mathematics

2 answers:

Blababa [14]2 years ago

6 0

Theres no picture or question

Veronika [31]2 years ago

5 0

Theres no question.

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Please help as fast as you can. This is due soon.

TEA [102]

Answer:

arc HK = 60

Step-by-step explanation:

angle HLK is an inscribed angle and arc HK is the arc it intercepts

an inscribed angle is equal to half the measure of the arc it intercepts

this can also be said as the intercepted arc is equal to twice the measure of the inscribed angle

Hence arc HK = 30 * 2 = 60

4 0

3 years ago

Read 2 more answers

1. Let f(x) = 5x + 3 and g(x) = 3x<br> What is (f + g)(x)?

mina [271]

Answer:

$8x + 3$

Step-by-step explanation:

In the picture.

I hope that it's a clear solution.

7 0

3 years ago

Solve (2x+3) (x+4)= 1

Digiron [165]

To solve, simply layout an equation.

(2x+3)(x+4)=1

Step 1: Simplify both sides of the equation.

2x2+11x+12=1

Step 2: Subtract 1 from both sides.

2x2+11x+12−1=1−1

2x2+11x+11=0

Step 3: Use quadratic formula with a=2, b=11, c=11.

x=-b±√b^2-4ac/2a

So, the answer for this problem is:

x=-11/4+1/4√33 or -11/4+-1/4√33

5 0

3 years ago

Mention the commutativity associative and distributive properties of rational number

evablogger [386]

Answer:

Associative property: a + (b + c), a – (b – c) ≠ (a – b) – c, a × (b × c) = (a × b) × c, and a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c

When a = ½ and b = ¾

Now, for checking a × b = b × a, consider LHS and RHS.

LHS = a × b = ½ × ¾ = ⅜

RHS = b × a = ¾ × ½ = ⅜

Thus, LHS = RHS (Hence proved)

Step-by-step explanation:

6 0

3 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0

3 years ago