PLEASE HELP!!

3rd question, I’m sorry guys but I was sick that’s why I didn’t attend school and the teacher said if I didn’t submit it by 5 ho

aniked [119]

Answer:

21. ΔUYZ Obtuse

22. ΔBCD Right or Rectangle

23. ΔADB Acute

24. ΔUXZ Acute

25. ΔUWZ Right or Rectangle

26. ΔUXY Equiangular and acute

Step-by-step explanation:

We need to classify each triangle by its angles.

Acute: They have 3 acute angles (less than 90 degrees).

Rectangle: The inner angle A is straight (90 degrees) and the other 2 angles are sharp. The sides that form the right angle are called legs (c and b), the other side hypotenuse.

Obtuse: The inner angle A is obtuse (more than 90 degrees). The other 2 angles are acute.

ΔUYZ Obtuse The inner angle Y is obtuse (more than 90°).

ΔBCD Right or Rectangle The inner angle C is straight (90°).

ΔADB Acute It has 3 acute angles (less than 90°).

ΔUXZ Acute It has 3 acute angles (less than 90°).

ΔUWZ Right or Rectangle The inner angle W is straight (90°).

ΔUXY Equiangular and acute Which means all the inner angles are equal (60°).

3 0

3 years ago

a box or laundry detergent is on sale for $6.50. If this price represents a 40% discount from the original price, what is the or

Usimov [2.4K]

The discount is $2.60. $6.50 + $2.60 = $9.10. $9.10 is the original price.

4 0

4 years ago

Read 2 more answers

Write a ratio and a percent for the shaded area.

nika2105 [10]

I’m pretty sure it’s 36/100 - 36% since there are 100 boxes and 36 of them are shaded

6 0

3 years ago

Eight cards are marked 3, 4, 5, 6, 7, 8, 9, and 10 such that each card has exactly one of these numbers. A card is picked withou

lawyer [7]

Answer: 3/8, 0.375, 37.5%

Step-by-step explanation:

Count the number of cards with odd numbers that are greater than 4. 5, 7, and 9 are odd numbers greater than 4. There are 3 cards with odd numbers greater than 4. Probability is found by dividing this by the total number of cards, 8. This gets you a fraction of 3/8.

6 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

4 years ago