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3 years ago

Can anyone help me do this

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

none of the others make sense

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<img src="https://tex.z-dn.net/?f=%5Csf%20-%286-4x%29" id="TexFormula1" title="\sf -(6-4x)" alt="\sf -(6-4x)" align="absmiddle"

Margaret [11]

Answer:

$- (6 - 4x) = - 6 + 4x$

8 0

3 years ago

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How do I evaluate the expression (8-4) x 2

ICE Princess25 [194]

Subtract (8-4) and you get 4 as the total and then times it by 2 which will give u 8, the answer is 8

3 0

2 years ago

A 400 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 18° angle with th

notka56 [123]

Answer:

left: 123.607 lb
right: 380.423 lb

Step-by-step explanation:

By balancing horizontal and vertical forces, we find the cable tensions to be ...

Ta = W·sin(b)/sin(a+b) . . . . . where W is the weight being held

Tb = W·sin(a)/sin(a+b)

Where Ta is the tension in the cable that makes an angle of 'a' with respect to the vertical, and Tb is the tension in the cable that makes an angle of 'b' with respect to the vertical.

The given angles are with respect to the ceiling, so the angles with respect to the vertical will be their compmements.

<h3>left cable (a)</h3>

angle 'a' is 90° -18° = 72°

angle 'b' is 90° -72° = 18°

a+b = 72° +18° = 90°

Ta = (400 lb)sin(18°)/sin(90°) = 123.607 lb

<h3>right cable (b)</h3>

Tb = (400 lb)sin(72°)/sin(90°) = 380.423 lb

_____

<em>Additional comment</em>

The nice expressions for cable tension come from the balance of forces.

vertical: Ta·cos(a) +Tb·cos(b) = W

horizontal: Ta·sin(a) = Tb·sin(b)

Solving the horizontal equation for Ta, we get ...

Ta = Tb·sin(b)/sin(a)

Substituting into the vertical equaiton gives ...

Tb·sin(b)cos(a)/sin(a) +Tb·cos(b) = W

Multiplying by sin(a) gives ...

Tb(sin(b)cos(a) +sin(a)cos(b)) = W·sin(a)

Using the trig identity for the sine of the sum of angles, we can rewrite this in the form shown above:

Tb = W·sin(a)/sin(a+b)

The problem is symmetrical with respect to 'a' and 'b', so the other tension is found by interchanging 'a' and 'b' in the equation:

Ta = W·sin(b)/sin(a+b)

4 0

2 years ago

''the product of a number and three is the same as the sum of that number and six''

Oduvanchick [21]

Answer:

Step-by-step explanation:

3 0

4 years ago

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Find the area of each shaded region each outer polygon is regular <br><br><br> See the picture

PSYCHO15rus [73]

Answer:

Area of the shaded region = 3.46 square units

Step-by-step explanation:

Measure of the interior angle of a regular polygon = $\frac{(n - 2)\times180}{n}$

From the picture attached,

Number of sides of the given polygon 'n' = 6

Interior angle (∠BAF) of the given polygon = $\frac{(6 - 2)\times 180}{6}$

= 120°

Measure of ∠BAC = $\frac{120}{4}$

= 30°

Now we apply sine rule in ΔAGB,

sin(30°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

= $\frac{BG}{AB}$

BG = AB[sin(30°)]

= $2\times \frac{1}{2}$

= 1

By applying cosine rule in ΔABG,

cos(30°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

= $\frac{AG}{AB}$

AG = AB[cos(30°)]

= $2\times \frac{\sqrt{3} }{2}$

= $\sqrt{3}$

AC = 2(AG)

= 2√3

Area of ΔABC = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(AC)(BG)$

= $\frac{1}{2}(2\sqrt{3} )(1)$

= $\sqrt{3}$

Area of the shaded region = Area of ΔABC + Area of ΔFED

= 2(Area of ΔABC)

= 2√3

= 3.46 square units

6 0

3 years ago