All categories

2 years ago

Which number has an 8 that is 1/10 the value of the 8 in 78?

Mathematics

1 answer:

eimsori [14]2 years ago

6 0

Step-by-step explanation:

Step one:

Firstly we need to determine the place value of 8 in the term 78

in the term, we have 7 tens and 8 units

hence the place value of 8 in 78 is the units

Step two:

Now 1/10 of 8 is given as

0.1*8= 0.8

According to the problem statement, we are expected to find a number that has an 8 that is the same as 0.8.

Say 5.8, 6.8, 9.8

we cannot say exactly, but any number that has an 8 that is one decimal away is an equivalent

You might be interested in

I am a 4 digit number larger than 2,500. No two of my digits are the same. The product of the thousands digit and the hundreds d

gulaghasi [49]

Answer:

After 5 minutes of hitting my head on a wall I found your answer... 3256

Step-by-step explanation:

4 0

2 years ago

Which points lie in Quadrant III? Choose all answers that are correct. A (–8, –2) B(–5, 4) C(–3, –9) D (4, –1) E (7, 0)

Studentka2010 [4]

Quadrant 3 will have a negative x and a negative y........so ur answers are :
A and C

4 0

2 years ago

Read 2 more answers

Which statements describe the functions? Check all

ruslelena [56]

Answer:

a,d,e

Step-by-step explanation:

edgenuity

6 0

2 years ago

Read 2 more answers

The formula for wind chill C (in degrees Fahrenheit) is given by C = 35.74 + 0.6215T - 35.75v^0.16 + 0.4275Tv^0.16 where v is th

Diano4ka-milaya [45]

Answer:

dC = 2.44

Relative error = 19%

Step-by-step explanation:

$C = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)$

Δv = 3 ; ΔT = 1 , v = 23, T = 8

Use differential Calculus

$dC = Cv.dv + Ct.dt$

$Cv = dC/dv = -5.72*v^(-0.84) + 0.0684*T*v^(-0.84)\\Ct = dC/dT = 0.6215 + 0.4275*v^(0.16)\\dC = modulus (Cv.dv) + modulus (CT.dT)\\dC = (-5.72*v^(-0.84) + 0.0684*T*v^(-0.84))* 3 + (0.6215 + 0.4275*v^(0.16))*1\\dC = 2.44$

Relative Error

dC / C @(T = 8, v=23) * 100 = 19 %

8 0

2 years ago

Use matrices and elementary row to solve the following system:

LiRa [457]

I assume the first equation is supposed to be

$5x-3y+2z=13$

and not

$5x-3x+2x=4x=13$

As an augmented matrix, this system is given by

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]$

Multiply through row 3 by 1/2:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]$

Add -1(row 2) to row 3:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]$

Multiply through row 3 by 1/5:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]$

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

$\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -3(row 2) to row 1:

$\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Multiply through row 1 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -2(row 1) to row 2:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]$

Multipy through row 2 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]$

The solution to the system is then

$\boxed{x=1,y=-2,z=1}$

5 0

2 years ago