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3 years ago

Describe the error. Please please help me. No links.

Mathematics

1 answer:

melisa1 [442]3 years ago

6 0

Triangle QRS is equal to triangle VWX by the ASA theorem. Not AAS

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Which statement is true?

muminat

The statement that is true is this one: The equation 3.5|6x – 2| = 3.5 has one solution

hope this helps!

5 0

3 years ago

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15x+12=45+4x whats x?

saul85 [17]

15x+12=45+4x ⇒15x - 4x = 45 - 12 ⇒ 11x = 33 ⇒ x = 3

4 0

3 years ago

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Brianna has $25 and a $10 gift certificate to a clothing store she buys two T-shirts for $9.97 each what are the questions you m

arlik [135]

Answer:

$15.06

Step-by-step explanation:

First you add $25.00 and $10.00 Then you multiply $9.97 by two and you would get $19.94 which you would then subtract $19.94 from $35

4 0

4 years ago

the surface area of a cuboid is 95cm² and its lateral surface area is 63cm². find the area of its base

wel

Answer:

The Area of the base is 16 cm² .

Step-by-step explanation:

Given as :

The surface area of the cuboid = x = 95 cm²

The lateral surface area of the cuboid = y = 63 cm²

Let The Area of the base = z cm²

Now, Let The length of cuboid = l cm

The breadth of cuboid = b cm

The height of cuboid = h cm

According to question

∵ The surface area of the cuboid = 2 ×(length × breadth + breadth × height + height × length)

Or, x = 2 ×(l × b + b × h + h × l)

Or, 95 = 2 ×(l × b + b × h + h × l)

Or, (l × b + b × h + h × l) = $\dfrac{95}{2}$ ....1

Similarly

∵lateral surface area of the cuboid = 2 ×(breadth × height + length × height)

Or, y = 2 ×(b × h + l × h)

Or, 2 ×(b × h + l × h) = 63

Or, (b × h + l × h) = $\dfrac{63}{2}$ ......2

Putting value of eq 2 into eq 1

so, (l × b + $\dfrac{63}{2}$ ) = $\dfrac{95}{2}$

Or, l × b = $\dfrac{95}{2}$ - $\dfrac{63}{2}$

Or, l × b = $\dfrac{95 - 63}{2}$

i.e l × b = $\dfrac{32}{2}$

so, l × b = 16

Now, Again

∵The Area of the base = ( length × breadth ) cm²

So, z = l × b

i.e z = 16 cm²

So, The Area of the base = z = 16 cm²

Hence,The Area of the base is 16 cm² . Answer

5 0

3 years ago

Solve for x -sin^2(x)=2cos(x)-2

Zolol [24]

We know that sin^2(x) = 1 - cos^2(x), so

2 - 2 cos^2(x) = 2 + cos(x)
cos(x) + 2 cos^2(x) = 0

One immediate possibility is that cos(x) = 0, in which case x = Pi/2 or 3pi/4. If cos(x) is not zero, then divide both sides by cos(x) to obtain

cos(x) = -1/2

which gives x = 2Pi/3 and x = 4Pi/3.

So your solutions are

Pi/2, 3Pi/4, 2Pi/3, and 4Pi/3

Hope this helps :D

6 0

3 years ago

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