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3 years ago

PLEASE HELP ASAP!! i will mark you brainliest

Mathematics

1 answer:

V125BC [204]3 years ago

4 0

Answer:

1. I am not sure about the first one. I tried everything and nothing I got from any of my work is looking right.

2. AB = 5

3. 8

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How do you solve addition equations using models?

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1. The networking organization you joined is throwing a party. You are in charge of buying the chips, which cost $2.50 per bag a

lapo4ka [179]

Answer:

Option d :18 bags of chips and 6 jars of salsa

Step-by-step explanation:

Given : The networking organization you joined is throwing a party. You are in charge of buying the chips, which cost $2.50 per bag and salsa, which costs $4 per jar. The chips & salsa budget you are given totals $60.

Inequality : $2.5x+4y\leq 60$

Solution :

x represents chips

y represents salsa

Option a: 10 bags of chips and 2 jars of salsa

so, x = 10 and y =2

Putting values in inequality

$2.5(10)+4(2)\leq 60$

$25+8\leq 60$

$33\leq 60$

Hence it is correct.

Option b : 20 bags of chips and 2 jars of salsa

so, x = 20 and y =2

Putting values in inequality

$2.5(20)+4(2)\leq 60$

$50+8\leq 60$

$58\leq 60$

Hence it is correct.

Option c : 14 bags of chips and 5 jars of salsa

so, x = 14 and y =5

Putting values in inequality

$2.5(14)+4(5)\leq 60$

$35+20\leq 60$

$55\leq 60$

Hence it is correct.

Option d :18 bags of chips and 6 jars of salsa

so, x = 18 and y =6

Putting values in inequality

$2.5(18)+4(6)\leq 60$

$45+24\leq 60$

$69\geq 60$

Hence it is not correct since it violates the inequality

4 0

3 years ago

Read 2 more answers

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