Ask question

Ask question

All categories

3 years ago

7

Mrs. Thomas bought a bed for $1,548 and three armchairs. The bed cost 4 times as much as one armchair. How much did Mrs. Thomas

spend altogether?

2 answers:

lakkis [162]3 years ago

8 0

1,548 multiplied times four equals 6,192

xxMikexx [17]3 years ago

8 0

Mrs. Thomas Spent 6,192 All Together.

You might be interested in

Frankie wants to build a path from one corner of his hard to the opposite corner his yard measures 20ft x 32ft what will be the

Papessa [141]

Answer:

37.7 feet

Step-by-step explanation:

Given that:

Measurements of Frankie's yard = 20ft x 32ft

He wants to build a path from one corner to another.

To find:

The length of path = ?

Solution:

Let us consider a rectangle $ABCD$ as shown in the attached image in the answer area.

We are given the two adjacent sides of a rectangle and we have to find diagonal of the rectangle.

To find the diagonal, we can use the Pythagorean Theorem.

We are given the base and perpendicular of the right angled triangle and we have to find the hypotenuse.

According to Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}$

$Diagonal = 20^2 + 32^2\\\Rightarrow Diagonal = 400 + 1024\\\Rightarrow Diagonal = \bold{37.7\ ft}$

4 0

3 years ago

What is the domain of y=log(x+2)

Andru [333]

Answer:

$\huge\boxed{x>-2\to x\in(-2;\ \infty)}$

Step-by-step explanation:

$y=\log(x+2)\\\\\bold{Domain:}\\\\x+2>0\qquad|\text{subtract 2 from both sides}\\\\x+2-2>0-2\\\\x>-2$

8 0

3 years ago

Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

7 0

3 years ago

Jada has a rectangular garden that is 10 meters long and 20 meters wide. (1 meter = 39.37 inches) how long is the garden measure

Rina8888 [55]

I think the answer is if you multiply. 10m X 20w= 200, now if you devide 39.37 in 200 is equal to 5.08in

4 0

4 years ago

A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of

Gwar [14]

$\bold{\huge{\underline{ Solution}}}$

<h3><u>Given </u><u>:</u><u>-</u></h3>

A marker in the center of the fairway is 150 yards away from the centre of the green
While standing on the marker and facing the green, the golfer turns 100° towards his ball
Then he peces off 30 yards to his ball

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>

<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>

Let assume that the distance between the golf ball and central of green is x

<u>Here</u><u>, </u>

Distance between marker and centre of green is 150 yards
<u>That </u><u>is</u><u>, </u>Height = 150 yards
For facing the green , The golfer turns 100° towards his ball
<u>That </u><u>is</u><u>, </u>Angle = 100°
The golfer peces off 30 yards to his ball
<u>That </u><u>is</u><u>, </u>Base = 30 yards

<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

$\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}$

Here, a = perpendicular height
b = base
c = hypotenuse
cos theta = Angle of cosine

<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

$\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}$

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}$

$\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}$

$\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}$

$\sf{ x^{2} = 22500 + 900 + 156.6}$

$\sf{ x^{2} = 23556.6}$

$\bold{ x = 153.48\: yards }$

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0

3 years ago