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oee [108]
3 years ago
12

In ΔABC, AC II DE and m∠1 = 55°. What is m∠4?

Mathematics
1 answer:
7nadin3 [17]3 years ago
4 0
M<1 = 55
m<4 = m<1 = 55 (corresponding angles)

hope it  helps
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At 11 p.M., it was 48°F. The temperature was 6° cooler by 7 a.M. By 11 a.M. The temperature was 48°F again. What integer represe
Anvisha [2.4K]

Answer:

6°F

Step-by-step explanation:

Given that:

Temperature at 11pm = 48°F

Temperature at 7am = 6° cooler

Temperature at 11am = 48°F

Number of degrees temperature changed from 7am to 11 am

Exact temperature at 7am = (temperature at 11 pm - 6°)

= 48°F - 6°F

= 42°F

Temperature change from 7am to 11am

Temperature at 11 am - temperature at 7am

48°F - 42°F

= 6°F

6 0
3 years ago
jaden made a pot of chili with 48 ounces of ground beef and 2 tablespoons of chili powder.he made another pot of chili with the
dexar [7]

Answer:

1/2 pound.

Step-by-step explanation:

Conversions: 48 oz = 3 lb

The ratio in the new pot is 3*2 tbs : 3 pounds , or just 6:3, which simplifies to 2:1. This means for every two tbs of powder, 1 lb of beef is used. This means that for every one tbs, 1/2 lb of beef is used.

8 0
3 years ago
Read 2 more answers
Given that √3  = 1.7321 , find correct to 3 places of decimals , the value of √192 - 1 / 2√48 - √75​
Flauer [41]

Step-by-step explanation:

the value of root three is given already so the question is asking you to round it of to the nearest 1000 which means there should be three numbers after the decimal point.

3 0
2 years ago
HELP
Aleks [24]

Answer:

c) 6

Step-by-step explanation:

This is a straight-forward application of the Law of Cosines:

... a² = b² + c² -2bc·cos(A)

... a² = 8² +11² -2·8·11·cos(32.2°) ≈ 64 +121 -176·0.8462

... a² ≈ 36.07000

... a ≈ 6.00583

The best choice is c) 6.

3 0
3 years ago
Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t
makkiz [27]

Answer:

Step-by-step explanation:

x + 2y + 3z = 12

z=\frac{12-x-2y}{3}

Volume = V=f(x,y) = xy(\frac{12-x-2y}{3} )

find partial derivatives using product rule

f_x =\frac{y}{3} (12-2x-2y)\\f_y = \frac{x}{3} (12-x-4y)

i.e.

Using maximum for partial derivatives, we equate first partial derivative to 0.

y=0 or x+y =6

x=0 or x+4y =12

Simplify to get y =2, x = 4

thus critical points are (4,2) (6,0) (0,3)

Of these D the II derivative test gives

D<0 only for (4,2)

Hence maximum volume is when x=4, y=2, z= 4/3

Max volume is = 4(2)(4/3) = 32/3

6 0
3 years ago
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