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xz_007 [3.2K]
3 years ago
11

The opening of the drinking straw was about 6 ? wide.​

Mathematics
1 answer:
dimulka [17.4K]3 years ago
8 0

Possible Answer:

Millimeter.

Step-by-step explanation:

Judging by the item in the question, a straw, millimeter is probably the correct measurement of the width of it.

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Joe was driving on the highway a car ahead of him was driving far below the speed limit so he decided to pass. In the first seco
spayn [35]

Answer:

537/100s.

Step-by-step explanation:

8 0
3 years ago
Please can someone help me on this I’m trying to revise for my test and I don’t understand this question
docker41 [41]

Answer:

∠ EAB = 85°

Step-by-step explanation:

∠ EBC and ∠ BCD are same- side interior angles and sum to 180° , so

∠ BCD = 180° - ∠ EBC = 180° - 148° = 32°

the sum of the 3 angles in Δ ADC = 180° , that is

∠ EDC + ∠ BCD + ∠ EAB = 180°

63° + 32° + ∠ EAB = 180°

95° + ∠ EAB = 180° ( subtract 95° from both sides )

∠ EAB = 85°

5 0
2 years ago
Read 2 more answers
Factor -3/4 out of 3/2x - 9/4y
Vikki [24]

Answer:

-\frac{3}{4} \times (-2x +3y)

Step-by-step explanation:

To factor out -3/4 out of that equation, multiply and divide by -3/4.

Say ;

=\frac{3}{2}x -\frac{9}{4}y

=(\frac{3}{2}x -\frac{9}{4}y) \times \frac{-\frac{3}{4} }{-\frac{3}{4}}

Now, rearranging and making necessary commutations.

=-\frac{3}{4} \times (\frac{3}{2}x -\frac{9}{4}y) \times \frac{1 }{-\frac{3}{4}}

=-\frac{3}{4} \times (-2x +3y)

Factoring -3/4 we get an equation without any fractions

= -2x + 3y

3 0
3 years ago
Find the value of k and x2<br> x^2+ 13x + k = 0, x1=-9
Anarel [89]

Given:

The quadratic equation is:

x^2+13x+k=0

x_1=-9

To find:

The value of k and x_1.

Solution:

We have,

x^2+13x+k=0                ...(i)

Putting x=-9, we get

(-9)^2+13(-9)+k=0

81-117+k=0

-36+k=0

k=36

Putting k=36 in (i), we get

x^2+13x+36=0

Splitting the middle term, we get

x^2+9x+4x+36=0

x(x+9)+4(x+9)=0

(x+9)(x+4)=0

x=-9,-4

Here, x_1=-9 and x_2=-4.

Therefore, the required values are k=36 and x_2=-4.

3 0
3 years ago
50 points
NISA [10]
1. Using your straightedge, draw a reference line, if one is not provided.
2. Copy the side of the square onto the reference line, starting at a point labeled A'.
3. Construct a perpendicular at point B' to the line through ab2.
4. Place your compass point at B', and copy the side of the square onto the perpendicular b'g. Label the end of the segment copy as point C.
5. With your compass still set at a span representing AB, place the compass point at C and swing an arc to the left.
6. Holding this same span, place the compass point at A' and swing an arc intersecting with the previous arc. Label the point of intersection as D.
7. Connect points A' to D, D to C, and C to B' to form a square.
4 0
3 years ago
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