Plz help and show work

Mrrafil [7]

The measure of the angle m<ABD is 90 degrees

<h3>Circle geometry</h3>

Circle geometry is the measure of angles within a circle. From te=he diagram shown, we can see that CD is tangent to circle A at point B and a line tangential to c circle is at 90 degrees

Since we are to find the measure of M<ABD, hence the required mesaure of the angle will be 90 degrees

Learn more on circle geometry here: brainly.com/question/24375372

7 0

2 years ago

Lunch is a hamburger which costs $4.95, an order of french fries for $1.85, and a drink for $1.05. Estimate the total cost to de

Colt1911 [192]

Answer:

$8.00

Step-by-step explanation:

Since it says estimate we are rounding. I'm assuming tot he nearest dollar.

$4.95 rounds to $5.00

$1.85 rounds to $2.00

$1.05 rounds to $1.00

Then add these up to see the total estimated cost. $8.00

6 0

3 years ago

HELP HELP HELP PLS If DF = 6x - 2 and FE = 10x - 18, then what is the length of DF?

lyudmila [28]

Answer: 22
6x-2=10x-18
6x-10x=-18+2
-4x=-16
x=4

6x-2
6(4)-2
24-2
=22

4 0

3 years ago

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The number of research articles in a prominent journal written by researchers during 1983–2003 can be approximated by U(t) = 4.1

Nimfa-mama [501]

Answer:

Step-by-step explanation:

From the information given:

We are to find; an expression for the total no. of articles written since 1983

∴

The total no. of articles written since 1983 $=\int \limts ^t_0 U(t) dt$

$= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]$

$=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

Therefore, the total number of articles written since 1983 is $=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

b. To find how many articles were being written from 1983 to 2003

i.e. t = 2003 - 1983 = 20

∴

Total articles written from 1983 to 2003 is $=\int \limts ^{20}_0 U(t) dt$

$= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]$

= 80.75 thousand articles

3 0

3 years ago

Solve for r: 3r+2-r=-4

earnstyle [38]

Answer:

r = -3

Step-by-step explanation:

3r + 2 - r = -4

2r + 2 = -4

2r = -6

r = -3

3 0

3 years ago

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