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dem82 [27]
2 years ago
6

FInd the length of DB and CD. Please help!

Mathematics
1 answer:
Lerok [7]2 years ago
3 0

Step-by-step explanation:

DB length = 5 1/4 + 2/3

=21/4 + 2/3

= 63/12 + 8/12 = 71/12

CD length = 5 1/4 -2 = 3 1/4

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May someone help me with this, it's fairly easy.
Likurg_2 [28]

Answer:

A and B are correct

Step-by-step explanation:

Student C is wrong because they broke up 3/4 into

2/2 + 1/2  

but 2/2 + 1/2 = 3/2, which is not equal to 3/4

7 0
2 years ago
a fruit delivers its fruit in two types of boxes: large and small. a delivery of 3 large boxes and 5 small boxes has a total wei
Masteriza [31]

Answer:

The weight of a small box  = 13.5 kg

The weight of 1 large box = 18.5 kg

Step-by-step explanation:

Let us assume the weight of a small box = m kg

And the weight of 1 large box  = n kg

Now, the weight of 5 small box = 5 x (weight of 1 small box) =  5 m

Also, the weight of 3 large box = 3 x (weight of 1 large box) =  3 n

Here,   3 large boxes +  5 small boxes =  of 123 kilograms

⇒ 5  m + 3 n = 123   .... (1)

Again, the weight of 2 small box = 2 x (weight of 1 small box) =  2 m

Also, the weight of 12 large box = 12 x (weight of 1 large box) =  12 n

Here,   12 large boxes +  2 small boxes =  249 kilograms

⇒ 2 m+ 12 n = 249   .... (2)

Now, solving both the given equations by ELIMINATION, we get:

5  m + 3 n = 123     x (2)

2 m+ 12 n = 249     x (-5)

we get the new set of equitation as:

10 m + 6 n = 246

- 10 m - 60 n =  -1245

Adding both equation, we get

-54 n = 999

or, n = 18.5

Now, 5  m + 3 n = 123

So, 5 m = 123  -3 (18.5)  = 123 - 55.5 =  67.5

⇒  m = 13.5

Hence,  the weight of a small box = m kg = 13.5 kg

And the weight of 1 large box = n kg  = 18.5 kg

8 0
3 years ago
Identify the vertex, focus, and directrix. y=1/8y^2
Brut [27]
If all the equations for the directrix are "x = " lines then this is a y^2 parabola.  The actual equation is x= \frac{1}{8}y^2.  The standard form for a positive sideways-opening parabola is  (y-k)^2=4p(x-h).  We know from the equation that the vertex of the parabola is at the origin, or else the translation would be reflected within the parenthesis in the equation.  Our equation has no parenthesis to indicate movement from the origin.  The vertex is (0, 0).  Got that out of the way.  That simplifies our standard form down to  y^2=4p(x).  Let's take a look at our equation now.  It is  x= \frac{1}{8}y^2.  We could rewrite it and make it a closer match to the standard form if we multiply both sides by 8 to get rid of the fraction.  That gives us an equation that looks like this:  y^2=8x.  That means that 4p = 8, and p = 2.  p is the distance that the focus and the directrix are from the vertex.  Since this is a positive parabola, it opens up to the right.  Which means, then, that the focus is to the right of the vertex, 2 units to be exact, and the directrix is 2 units to the left of the vertex.  The formula for the focus is (h + p, k).  Our h is 0, our k is 0 and our p is 2, so the coordinates of the focus are (2, 0).  Going 2 units to the left of the origin then puts our directrix at the line x = -2.  Your choice then as your answer is b. 
6 0
3 years ago
HELP PLLSS<br> 92;69<br> work included!
Nata [24]

Answer:

92/69 cannot be simplified

Step-by-step explanation:

8 0
3 years ago
Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4
weqwewe [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x = 3

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \:  log_{2}(x - 1)  log_{2}(x + 5)  = 4

\qquad \tt \rightarrow \:  log_{2} \{(x - 1)(x + 5) \} = 4

[ log (x) + log (y) = log (xy) ]

\qquad \tt \rightarrow \: ( x - 1)(x + 5) =  {2}^{4}

\qquad \tt \rightarrow \:  {x}^{2}  + 5x - x - 5 =  16

\qquad \tt \rightarrow \:  {x}^{2}  + 4x - 5 - 16 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 4x -21 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 7x - 3x - 21 = 0

\qquad \tt \rightarrow \:  x(x + 7) - 3(x + 7) = 0

\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0

\qquad \tt \rightarrow \: x =  - 7 \:  \: or \:  \: x = 3

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

\qquad \tt \rightarrow \: x = 3

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
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