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Whitepunk [10]
3 years ago
13

Are these correct? If not please help me write the correct answer.​

Mathematics
2 answers:
Llana [10]3 years ago
8 0

Answer:

i = correct

ii = correct

iii = correct

iv = correct

4

i = 16/81 => 0.1975308642

ii= 1/64

Keith_Richards [23]3 years ago
4 0

Answer:

Mostly correct

For Q3, you are doing great, you remembered that negative numbers can happen if the exponent is an odd number.

For Q4, I think you were confused with the fractions,

4(i)

(\frac{2}{3} )^{4} = (\frac{2^{4}}{3^{4}}) = (\frac{16}{81}) =0.198

4(ii) is also wrong but I'll let you try to fix it yourself following what I given you for 4(i). You should get 1/64

Hope this helps!

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Angelo, Brandon, and Carl work in the same office. Angelo’s age is 4 years more than twice Carl’s age. Brandon is 5 years younge
NNADVOKAT [17]

Answer:

Angelo is 66 Brandon is 26 Carl is 31

Step-by-step explanation:

Variables a=Angelo b=Brandon c=Carl

Equation (1/3)(a+b+c) = 41

7 0
3 years ago
Please please help me
Anarel [89]

so you are going to use the Pythagorean theorem on the smaller triangle to find the connecting side's length

{5}^{2}  = {3}^{2} ( {b}^{2} ) \\ b = 4

now you can find the bigger triangle's hypotenuse

{6}^{2}  +  {4}^{2}  =  {c}^{2}  \\

c=4

8 0
3 years ago
A group of 32 players forms 4 volleyball teams if there are 96 players how many teams can be formed
kotegsom [21]

Answer:

12 teams

Step-by-step explanation:

We know that 38 players can fit into 4 groups, 8 times ...

32 divide 8 = 4

We have the number with 96, and we need to find the groups

we do the same as done with the first one...

because 32 divided by 8 is 4

we do

96 divide 8 = 12

aswell

<h2><u><em>hope this helped : )</em></u></h2>
6 0
3 years ago
Read 2 more answers
From a point 100 m from a building the angles of elevation of the top and bottom of a flagpole atop a building are 54.5 degrees
AnnZ [28]

Answer:60ft

Step-by-step explanation:The height of the flagpole is approximately  

60

feet.

Explanation:

Always try to draw a diagram.

enter image source here

We know that there is a right angle between the ground and the building. Therefore, we can use the 3 basic trig ratios instead of the sine or cosine law to solve this problem.

Since the angle in the corner of the larger right triangle measures  

42

˚

, the top angle in this triangle measures  

180

˚

−

90

˚

−

42

˚

=

48

˚

.

By basic trig ratios, we can find the height of the building with the flag pole on top, call it  

H

.

tan

42

˚

1

=

H

500

H

=

500

tan

42

˚

I would keep it in exact form until the last step.

We now devise an expression for the height of the building (without the flag pole). Call it  

a

tan

38

˚

1

=

a

500

a

=

500

tan

38

˚

We can now state that

h

=

H

−

a

h

=

500

tan

42

˚

−

500

tan

38

˚

h

≈

59.559

≈

60

feet

Hopefully this helps!

Answer link

EET-AP

Apr 10, 2017

The flagpole is  

60

f

t

in height to the nearest foot.

Explanation:

1) The flagpole is on top of a building.

2)Angles of elevation both measured from point  

500

f

t

from building

3) Angle of elevation to the top of building is  

38

d

e

g

4) Angle of elevation to the top of flagpole is  

42

d

e

g

The information above will provide us with two right angle triangles, one smaller one inside a larger one.

Both will have a base of  

500

f

t

.

The smaller triangle will have a base angle  

β

of  

38

deg opposite the  

90

deg, and the larger triangle will have a base angle  

β

of  

42

deg.

From this information we can find the heights of the building and the building + pole using the definition of the tangent of the two base angles  

β

:

tan

(

β

)

=

o

p

p

a

d

j

where the  

o

p

p

is the height and the  

a

d

j

is the  

500

f

t

o

p

p

(

b

u

i

l

d

)

=

tan

(

38

)

⋅

(

500

f

t

)

=

390.6

f

t

=

height of building

o

p

p

(

f

l

a

g

)

=

tan

(

42

)

⋅

(

500

f

t

)

=

450.2

f

t

=

height of building + pole

Then to the nearest foot the height of the flagpole is:

450.2

f

t

−

390.6

f

t

=

60

f

t

4 0
3 years ago
Consider the 3 × 3 matrix A =   0 1 k 2 k −6 2 7 4  . For what values of k is matrix A invertible? (a) k ∈ R (b) all real k
Vanyuwa [196]

Answer:

b)

Step-by-step explanation:

A is invertible if and only if det(A)≠0. Let's compute the determinant of A and find the values k for which it is nonzero.

Using Sarrus's rule, we obtain that

\det A=0(k)(4)+1(-6)(2)+k(2)(7)-2(k)(k)-7(-6)(0)-4(2)(1)=-12+14k-2k^2-8=-2k^2+14k-20=-2(k^2-7k+10)=2(k-5)(k-2)

Note that the determinant is a quadratic equation on k, which can be factored as above.

Now the determinant is only zero if k=5 or k=2 (the zeroes of the quadratic polynomial). Therefore, if k≠2,5 the determinant is nonzero so A is invertible.

4 0
3 years ago
Read 2 more answers
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