How do you get the midpoint of a line segment? using shortcuts

Inverse of A=300e^-0.013t

AnnZ [28]

Answer:

$A^{-1}(t)=-\frac{ln(\frac{t}{300})}{0.013}$

Step-by-step explanation:

we have

$A=300e^{-0.013t}$

Exchange the variables A for t and t for A

$t=300e^{-0.013A}$

Isolate the variable A

$\frac{t}{300}=e^{-0.013A}$

Apply ln (natural logarithm) both sides

$ln(\frac{t}{300})=ln[e^{-0.013A}]$

$ln(\frac{t}{300})=(-0.013A)ln(e)$

Remember that

$ln(e)=1$

$ln(\frac{t}{300})=(-0.013A)$

$A=-\frac{ln(\frac{t}{300})}{0.013}$

Let

$A^{-1}(t)=A$

so

$A^{-1}(t)=-\frac{ln(\frac{t}{300})}{0.013}$

3 0

3 years ago

!!!!!PLEASE HELP!!!!! *30 POINTS PLEASE*

Lelechka [254]

Answer:

Part 1)

Part a) Yes, triangle ABC is congruent with triangle DEC by SAS

Part b) No, triangle ABC is not congruent with triangle DCE

Part 2) No, is not enough information

Part 3) Yes, triangle RST is congruent with triangle VUT by ASA

Part 4) Yes, triangle DEF is congruent with triangle DHG by SAS

Step-by-step explanation:

Part 1) we have two cases

Part a) Is triangle ABC and DEC congruent?

Yes, triangle ABC is congruent with triangle DEC by SAS

we know that

The Side Angle Side postulate ( SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent

In this problem

AC≅DE

BC≅EC

and

∠ACB≅∠DCE ---> by vertical angles

Part b) Is triangle ABC and DCE congruent?

we know that

If two triangles are congruent, then its corresponding sides and its corresponding angles are congruent

In this problem

the corresponding sides are not congruent

because

AB is not congruent with DC

Part 2) we have that

No, is not enough information

If AC≅XZ then the triangles ABC and XYZ will be congruent by SSS

or

If ∠ABC≅∠XYZ then the triangles ABC and XYZ will be congruent by SAS

Part 3) we have that

Yes, triangle RST is congruent with triangle VUT by ASA

we know that

If any two angles and the included side are the same in both triangles, then the triangles are congruent by Angle -Side-Angle (ASA) postulate

In this problem

ST≅UT

∠RST≅∠VUT ----> is a right angle

∠RTS≅∠VTU ---->by vertical angles

so

Two angles and the included side are the same in both triangles

therefore

triangles RST and VUT are congruent by ASA

Part 4) we have that

Yes, triangle DEF is congruent with triangle DHG by SAS

we know that

The Side Angle Side postulate ( SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent

In this problem

DE≅DH

DF≅DG

and

∠EDF≅∠HDG ---> by vertical angles

7 0

3 years ago

If the equation y=mx+b is used to model a quantity y as a function of the quantity x, why is b considered to be the starting val

blagie [28]

Answer:

This is because even when the quality of the variable x is 0, the value of b still exists and is the starting point for y.

Step-by-step explanation:

To prove this, we start with the base form of the equation and input 0 for x.

y = mx + b

y = m(0) + b

y = 0 + b

y = b

3 0

4 years ago

Charlie has 5 times as many stamps as Ryan. They have 1,608 stamps in all. How many more stamps does CHarlie have than Ryan?

Leokris [45]

So,
The ratio of Ryan to Charlie is 1:5
Ryan=x
Charlie=5x
5x+x=1608
6x=1608
Divide both sides by 6
1608/6=268
x=268
Ryan has 268 stamps.
Charlie has 5(268) 1340 stamps.
1340+268=1608
1340-268=1072.
Charlie has 1072 more stickers than Ryan.

Hope I helped!

4 0

3 years ago

Find the equation for the line that passes through the points and . State your

gavmur [86]

Answer:

y = 2x.

Step-by-step explanation:

The slope = (6 - (-4))/(3 - (-2))

= (6 + 4)/(3+2)

= 10/5

= 2.

Using this and the point (3, 6) we have:

y - 6 = 2(x - 3)

y = 2x - 6 + 6

y = 2x.

3 0

3 years ago