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Olegator [25]
3 years ago
5

Slimplify 14^0+2^7+15^1=

Mathematics
1 answer:
Leni [432]3 years ago
5 0

Answer:

1+128+15

Step-by-step explanation:

14^0 = 1

2^7 = 128

15^1 = 1

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Help please! ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!
solmaris [256]

Answer:

x= 1/2

and y=3

Step-by-step explanation:

Let's solve your system by substitution.

y=10x−2;2x+y=4

Step: Solvey=10x−2for y:

y=10x−2

Step: Substitute10x−2foryin2x+y=4:

2x+y=4

2x+10x−2=4

12x−2=4(Simplify both sides of the equation)

12x−2+2=4+2(Add 2 to both sides)

12x=6

12x

12

=

6

12

(Divide both sides by 12)

x=

1

2

Step: Substitute

1

2

forxiny=10x−2:

y=10x−2

y=10(

1

2

)−2

y=3(Simplify both sides of the equation)

Answer:

x= 1 /2

and y

3 0
3 years ago
Land in downtown Columbia is valued at $10 a square foot. What is the value of a triangular lot with sides of lengths 119, 147,
stiks02 [169]

Answer:

$87,461

Step-by-step explanation:

Given that the dimensions or sides of lengths of the triangle are 119, 147, and 190 ft

where S is the semi perimeter of the triangle, that is, s = (a + b + c)/2.

S = (119 + 147 + 190) / 2 = 456/ 2 = 228

Using Heron's formula which gives the area in terms of the three sides of the triangle

= √s(s – a)(s – b)(s – c)

Therefore we have = √228 (228 - 119)(228 - 147)(228 - 190)

=> √228 (109)(81)(38)

= √228(335502)

=√76494456

= 8746.1109071 * $10

= 87461.109071

≈$87,461

Hence, the value of a triangular lot with sides of lengths 119, 147, and 190 ft is $87,461.

6 0
3 years ago
WILL MARK BRANLIEST!! part a: Abe rented a bike at $36 for five days. If he rents the same bike for eight days, he has to pay a
AVprozaik [17]
<h3>Hello i hope you having a good day :) Question : Part a: Abe rented a bike at $36 for five days. If he rents the same bike for eight days, he has to pay a total rent of $48.</h3><h3>So we can set the fixed free of the renting that would be (f) and the dynamic free as (d) per the day then we have F + 5d = 36 F +  8d = 48 there for we have f = 16 and d = 4 so it would be  y = 16 + 4x.</h3><h3>Hopefully this help you.</h3>
5 0
3 years ago
Simply using properties of exponents
Luba_88 [7]

Answer:

Step-by-step explanation:

\frac{3^{2}x^{2}y^{2}}{3x^{\frac{1}{2}}}=3^{2-1}x^{2-\frac{1}{2}}y^{2}\\\\=3x^{3/2}y^{2}

5 0
3 years ago
An airplane flying at an altitude of 37,000 feet is still A horizontal distance of 100 miles (100 miles = 5280 ft) from the airp
vovikov84 [41]

Given :

An airplane flying at an altitude of 37,000 feet is still A horizontal distance of 100 miles (100 miles = 5280 ft) from the airport.

To Find :

What angle of depression does the airplane need to use to reach the runway?

Solution :

Let, angle of depression is \theta .

So,

tan\ \theta = \dfrac{ Altitude \ in \ which \ plane \ is \ flying}{Horizontal \ Distance}\\\\tan \ \theta = \dfrac{37000}{5280}\\\\tan \  \theta = 7.00\\\\\theta = tan^{-1} 7\\\\\theta = 81.87^o

Hence, this is the required solution.

4 0
3 years ago
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