2

Solve for w 6w + 4w - 9 + 6 = 9w - 7+ 6

Dmitry_Shevchenko [17]

Answer:

w = 2

Step-by-step explanation:

6w + 4w - 9 + 6 = 9w - 7+ 6

6w + 4w - 9w = - 7 + 6 + 9 - 6

w = 2

4 0

2 years ago

Leslie and Daniel plan to put an above-ground pool in the backyard of their new home. The backyard is rectangular, 30 feet by 50

Brut [27]

1047.84ft² is not covered by the pool.
Find the area of the yard covered by pull using the area of a circle formula (the height is irrelevant in this case). If the diameter of the pool is 24 feet, its radius is 12 (half of the diameter)

A = 3.14r^2
A =3.14(144)
A = 452.16 ft²

Subtract the area of the pool from the area of the yard to get the area of the yard that is not covered by the pool. If the dimensions of the yard are 30ft by 50ft, you multiply them to get the area: 1500ft²

Total yard area: 1,500ft²
Area of yard without pool: 1,500ft² - 452.16ft² = 1047.84ft²

4 0

3 years ago

Solve: 1/4x - 5 = 3/4x - 12

ivanzaharov [21]

The answer to the math problem it X=14 hope this helps

7 0

2 years ago

Read 2 more answers

Plz hurry!!!! thank you!!!!

Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = $\frac{115}{2}$ = 57.5°

Now, tan(57.5°) = $\frac{OS}{SN}$

⇒ 1.5697 = $\frac{R}{SN}$

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = $\frac{NQ}{QM}$

⇒ tan65° = $\frac{NQ}{QM}$

⇒ QM = $\frac{2R}{2.1445}$

QM = 0.9326 R .

⇒ MT = MQ + QT

= 0.9326 R + 0.637 R (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = ( $\frac{NP + MK}{2}$ ) × (ST)

= ( $\frac{1.274 R + 3.1392 R}{2}$ ) × 2 R

= 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0

2 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

2 years ago